We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

Create An Account

All Calculus 2 Resources

9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #78 : Series In Calculus

$\\ \textup{Use the ratio test to find out if the following series is convergent:}\\ \sum_{n = 1}^{\infty} \frac{10^n}{3^{2n}}\newline note: \hspace f(n) = \frac{10^n}{3^{2n}}$

Possible Answers:

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{1}{3}$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{10}{9}$

Correct answer:

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{10}{9}$

Explanation:

$\newline \textnormal{Determine the convergence of the series based on the limits.}\newline Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} < 1 \newline Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} > 1 \newline Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1\newline \textnormal{Solution:} \newline\textnormal{1. Ignore constants and simplify the equation (canceling out what you can)}\newline\textnormal{2. Once the equation is simplified, take} \lim_{n\rightarrow \infty}$

$\frac{\frac{10^{\left(n+1\right)}}{3^{2\left(n+1\right)}}}{\frac{10^n}{3^{2n}}} = \frac{3^{2n}\cdot \:10^{n+1}}{3^{2\left(n+1\right)}\cdot \:10^n} = \frac{10}{9}$

$\lim_{n\rightarrow \infty} \frac{10}{9} = \frac{10}{9}$

Report an Error

Example Question #79 : Series In Calculus

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{50n!}{n^{4}ln(n)} \newline note: \hspace f(n) = \frac{50n!}{n^{4}ln(n)}$

Possible Answers:

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{1}{4}$

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 4$

Correct answer:

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

Explanation:

$\newline \textnormal{Determine the convergence of the series based on the limits.}\newline Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} < 1 \newline Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} > 1 \newline Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1\newline\textnormal{Solution:} \newline\textnormal{1. Ignore constants and simplify the equation (canceling out what you can)}\newline\textnormal{2. Once the equation is simplified, take} \lim_{n\rightarrow \infty}$

$\frac{\frac{\left(n+1\right)!}{\left(n+1\right)^4\ln \left(\left(n+1\right)\right)}}{\frac{n!}{n^4\ln \left(n\right)}} = \frac{(n+1)!n^4ln(n)}{n!(n+1)^4ln(n+1)} = \frac{n^4ln(n)}{(n+1)^3ln(n+1)}$

$\lim_{n\rightarrow \infty} \frac{n^4ln(n)}{(n+1)^3ln(n+1)} = \infty$

Report an Error

Example Question #80 : Series In Calculus

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{70n!}{n^{8}ln(n)}\newline note: \hspace f(n) = \frac{70n!}{n^{8}ln(n)}$

Possible Answers:

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{1}{8}$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 8$

Correct answer:

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

Explanation:

$\newline \textnormal{Determine the convergence of the series based on the limits.}\newline Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} < 1 \newline Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} > 1 \newline Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1\newline\textnormal{Solution:} \newline\textnormal{1. Ignore constants and simplify the equation (canceling out what you can)}\newline\textnormal{2. Once the equation is simplified, take} \lim_{n\rightarrow \infty}$

$\frac{\frac{\left(n+1\right)!}{\left(n+1\right)^8\ln \left(\left(n+1\right)\right)}}{\frac{n!}{n^8\ln \left(n\right)}} = \frac{(n+1)!n^8ln(n)}{n!(n+1)^8ln(n+1)} = \frac{n^8ln(n)}{(n+1)^7ln(n+1)}$

$\lim_{n\rightarrow \infty} \frac{n^8ln(n)}{(n+1)^7ln(n+1)} = \infty$

Report an Error

Example Question #81 : Series In Calculus

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{35n!}{17n^{10}ln(n)}\newline note: \hspace f(n) = \frac{35n!}{17n^{10}ln(n)}$

Possible Answers:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{1}{10}$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 10$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

Correct answer:

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

Explanation:

$\newline \textnormal{Determine the convergence of the series based on the limits.}\newline Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} < 1 \newline Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} > 1 \newline Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1\newline\textnormal{Solution:} \newline\textnormal{1. Ignore constants and simplify the equation (canceling out what you can)}\newline\textnormal{2. Once the equation is simplified, take} \lim_{n\rightarrow \infty}$

$\frac{\frac{\left(n+1\right)!}{\left(n+1\right)^{10}\ln \left(\left(n+1\right)\right)}}{\frac{n!}{n^{10}\ln \left(n\right)}} = \frac{(n+1)!n^{10}ln(n)}{n!(n+1)^{10}ln(n+1)} = \frac{n^{10}ln(n)}{(n+1)^{9}ln(n+1)}$

$\lim_{n\rightarrow \infty} \frac{n^{10}ln(n)}{(n+1)^{9}ln(n+1)}= \infty$

Report an Error

Example Question #82 : Series In Calculus

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{20n!}{n^{6}ln(n)}\newline note: \hspace f(n) = \frac{20n!}{n^{6}ln(n)}$

Possible Answers:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{1}{6}$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 6$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

Correct answer:

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

Explanation:

$\newline \textnormal{Determine the convergence of the series based on the limits.}\newline Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} < 1 \newline Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} > 1 \newline Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1\newline\textnormal{Solution:} \newline\textnormal{1. Ignore constants and simplify the equation (canceling out what you can)}\newline\textnormal{2. Once the equation is simplified, take} \lim_{n\rightarrow \infty}$

$\frac{\frac{\left(n+1\right)!}{\left(n+1\right)^6\ln \left(\left(n+1\right)\right)}}{\frac{n!}{n^6\ln \left(n\right)}} = \frac{(n+1)!n^6ln(n)}{n!(n+1)^6ln(n+1)} = \frac{(n+1)n^6ln(n)}{(n+1)^6ln(n+1)}$

$\lim_{n\rightarrow \infty} \frac{(n+1)n^6ln(n)}{(n+1)^6ln(n+1)}= \infty$

Report an Error

Example Question #83 : Series In Calculus

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{51*3^{n}n!}{2*n^{n}*e^{n}}\newline note: \hspace f(n) = \frac{51*3^{n}n!}{2*n^{n}*e^{n}}$

Possible Answers:

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{e^2}{3}$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{3}{e^2 }$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

Correct answer:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{3}{e^2 }$

Explanation:

$\newline \textnormal{Determine the convergence of the series based on the limits.}\newline Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} < 1 \newline Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} > 1 \newline Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1\newline\textnormal{Solution:} \newline\textnormal{1. Ignore constants and simplify the equation (canceling out what you can)}\newline\textnormal{2. Once the equation is simplified, take} \lim_{n\rightarrow \infty}$

$\frac{\frac{3^{\left(n+1\right)}\left(n+1\right)!}{\left(n+1\right)^{\left(n+1\right)}e^{n+1}}}{\frac{3^nn!}{n^ne^n}} = \frac{3^{n+1}(n+1)!n^{n}e^{n}}{3^{n}n!(n+1)^{n+1}e^{n+1}} = \frac{3n^n}{(n+1)^{n}e}$

$\lim_{n\rightarrow \infty} \frac{3n^n}{(n+1)^{n}e} = \frac{3}{e^2}$

$\newline \textnormal {Note: By L'Hopital's rule} \newline\lim_{n\rightarrow \infty} \frac{n^{n}}{(n+1)^{n}} = \frac{1}{e}$

Report an Error

Example Question #84 : Series In Calculus

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{11*4^{n}n!}{16*n^{n}*e^{n}}\newline note: \hspace f(n) = \frac{11*4^{n}n!}{16*n^{n}*e^{n}}$

Possible Answers:

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{4}{e^2}$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{e^2}{4}$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

Correct answer:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{4}{e^2}$

Explanation:

$\newline \textnormal{Determine the convergence of the series based on the limits.}\newline Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} < 1 \newline Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} > 1 \newline Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1\newline\textnormal{Solution:} \newline\textnormal{1. Ignore constants and simplify the equation (canceling out what you can)}\newline\textnormal{2. Once the equation is simplified, take} \lim_{n\rightarrow \infty}$

$\frac{\frac{4^{\left(n+1\right)}\left(n+1\right)!}{\left(n+1\right)^{\left(n+1\right)}e^{n+1}}}{\frac{4^nn!}{n^ne^n}} = \frac{4^{n+1}(n+1)!e^{n}n^{n})}{4^{n}n!e^{n+1}(n+1)^{n+1}} = \frac{4n^n}{e(n+1)^n}$

$\lim_{n\rightarrow \infty} \frac{4n^n}{e^n(n+1)^n} = \frac{4}{e^2}$

$\newline \textnormal {Note: By L'Hopital's rule} \newline\lim_{n\rightarrow \infty} \frac{n^{n}}{(n+1)^{n}} = \frac{1}{e}$

Report an Error

Example Question #85 : Series In Calculus

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{25*5^{n}*n!}{n^{n}e^n}\newline note: \hspace f(n) = \frac{25*5^{n}*n!}{n^{n}e^n}$

Possible Answers:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{5}{e^2}$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{e^2}{5}$

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

Correct answer:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{5}{e^2}$

Explanation:

$\newline \textnormal{Determine the convergence of the series based on the limits.}\newline Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} < 1 \newline Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} > 1 \newline Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1\newline\textnormal{Solution:} \newline\textnormal{1. Ignore constants and simplify the equation (canceling out what you can)}\newline\textnormal{2. Once the equation is simplified, take} \lim_{n\rightarrow \infty}$

$\frac{\frac{5^{\left(n+1\right)}\left(n+1\right)!}{\left(n+1\right)^{\left(n+1\right)}e^{n+1}}}{\frac{5^nn!}{n^ne^n}} = \frac{5^{n+1}(n+1)!n^ne^n}{5^{n}n!(n+1)^{n+1}e^{n+1}} = \frac{5n^n}{(n+1)^ne}$

$\lim_{n\rightarrow \infty} \frac{5n^n}{(n+1)^ne} = \frac{5}{e^2}$

$\newline \textnormal {Note: By L'Hopital's rule} \newline\lim_{n\rightarrow \infty} \frac{n^{n}}{(n+1)^{n}} = \frac{1}{e}$

Report an Error

Example Question #86 : Series In Calculus

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{36*6^{n}*n!}{n^{n}e^n}\newline note: \hspace f(n) = \frac{36*6^{n}*n!}{n^{n}e^n}$

Possible Answers:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{e^2}{6}$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{6}{e^2}$

Correct answer:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{6}{e^2}$

Explanation:

$\newline \textnormal{Determine the convergence of the series based on the limits.}\newline Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} < 1 \newline Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} > 1 \newline Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1\newline\textnormal{Solution:} \newline\textnormal{1. Ignore constants and simplify the equation (canceling out what you can)}\newline\textnormal{2. Once the equation is simplified, take} \lim_{n\rightarrow \infty}$

$\frac{\frac{6^{\left(n+1\right)}\left(n+1\right)!}{\left(n+1\right)^{\left(n+1\right)}e^{n+1}}}{\frac{6^nn!}{n^ne^n}} = \frac{6^{n+1}(n+1)!n^ne^n}{6^nn!(n+1)^{n+1}e^{n+1}} = \frac{6n^n}{(n+1)^ne}$

$\lim_{n\rightarrow \infty}\frac{6n^{n}}{(n+1)^{n}e} = \frac{6}{e^2}$

$\newline \textnormal {Note: By L'Hopital's rule} \newline\lim_{n\rightarrow \infty} \frac{n^{n}}{(n+1)^{n}} = \frac{1}{e}$

Report an Error

Example Question #87 : Series In Calculus

$\textnormal{Use the ratio test to find out if the following series is convergent:} \newline\sum_{n = 1}^{\infty} \frac{7^{n}*n!}{21n^{n}}\newline note: \hspace f(n) = \frac{7^{n}*n!}{21n^{n}}$

Possible Answers:

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{7}{e}$

$nconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{e}{7}$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

Correct answer:

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{7}{e}$

Explanation:

$\newline \textnormal{Determine the convergence of the series based on the limits.}\newline Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} < 1 \newline Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} > 1 \newline Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1\newline\textnormal{Solution:} \newline\textnormal{1. Ignore constants and simplify the equation (canceling out what you can)}\newline\textnormal{2. Once the equation is simplified, take} \lim_{n\rightarrow \infty}$

$\frac{7^{\left(n+1\right)}\frac{\left(n+1\right)!}{\left(n+1\right)^{\left(n+1\right)}}}{7^n\frac{n!}{n^n}}= \frac{7n^{n}}{(n+1)^{n}}$

$\\\lim_{n\rightarrow \infty}\frac{7n^{n}}{(n+1)^{n}} = \frac{7}{e} \\ \newline \textnormal {Note: By L'Hopital's rule} \newline\lim_{n\rightarrow \infty} \frac{n^{n}}{(n+1)^{n}} = \frac{1}{e}$

Report an Error

View Calculus 2 Tutors

Yoonsik
Certified Tutor

Seoul National University, Bachelor of Science, Physics. University of Pennsylvania, Doctor of Philosophy, Atomic and Molecul...

View Calculus 2 Tutors

Rod
Certified Tutor

University of Minnesota-Twin Cities, Bachelor of Science, Mathematics Teacher Education. University of Minnesota-Twin Cities,...

View Calculus 2 Tutors

Chase
Certified Tutor

Western Governors University, Bachelor of Science, Mathematics.

All Calculus 2 Resources

9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

GMAT Tutors in Phoenix, ISEE Tutors in Chicago, GRE Tutors in Dallas Fort Worth, ISEE Tutors in Seattle, SSAT Tutors in Houston, English Tutors in Dallas Fort Worth, LSAT Tutors in Washington DC, Computer Science Tutors in Dallas Fort Worth, Math Tutors in Phoenix, Reading Tutors in Seattle

Popular Courses & Classes

MCAT Courses & Classes in Chicago, GRE Courses & Classes in Houston, GRE Courses & Classes in Denver, SAT Courses & Classes in Chicago, GRE Courses & Classes in New York City, GRE Courses & Classes in Chicago, SAT Courses & Classes in Boston, MCAT Courses & Classes in Miami, ACT Courses & Classes in Atlanta, MCAT Courses & Classes in Phoenix

Popular Test Prep

LSAT Test Prep in Houston, ISEE Test Prep in Los Angeles, GRE Test Prep in New York City, ISEE Test Prep in Miami, GMAT Test Prep in New York City, MCAT Test Prep in Houston, LSAT Test Prep in Dallas Fort Worth, SAT Test Prep in Houston, GMAT Test Prep in Houston, MCAT Test Prep in Phoenix

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #78 : Series In Calculus

Example Question #79 : Series In Calculus

Example Question #80 : Series In Calculus

Example Question #81 : Series In Calculus

Example Question #82 : Series In Calculus

Example Question #83 : Series In Calculus

Example Question #84 : Series In Calculus

Example Question #85 : Series In Calculus

Example Question #86 : Series In Calculus

Example Question #87 : Series In Calculus

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: