We can find the radius of convergence of

\(\displaystyle \sum_{n=0}^{\infty} \frac{x^{5n}}{3^n}\)

using the ratio test:

\(\displaystyle \lim_{n\to \infty} \left|\frac{a_{n+1}}{a_n} \right |=\lim_{n\to \infty} \left|\frac{x^{5n+5}}{3^{n+1}}\frac{3^n}{x^{5n}} \right |=\lim_{n\to \infty} \left|\frac{x^{5}}{3} \right |=\left|\frac{x^5}{3}\right|< 1\)

which means that \(\displaystyle |x|< 3^{\frac{1}{5}}\)

So that \(\displaystyle R=3^{\frac{1}{5}}\) is the radius of convergence.

To determine whether the series given converges or diverges, we must use the Ratio Test, which states that for 

\(\displaystyle L=\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_{n}}\right |\), if \(\displaystyle L< 1\), then series is (absolutely) convergent; if \(\displaystyle L=1\), then the series may be (absolutely) convergent, divergent, or conditionally convergent; and if \(\displaystyle L>1\), then the series is divergent. 

So, we must evaluate the limit as given by the above formula:

\(\displaystyle L=\lim_{n\rightarrow \infty } \frac{(n+1)!}{3^{n+3}}\cdot \frac{3^{n+2}}{(n)!}=\lim_{n\rightarrow \infty } \frac{n+1}{3}=\infty >1\)

Thus, the series is divergent. 

(The limit was solved by \(\displaystyle \lim_{n\rightarrow \infty } \frac{n}{n}(\frac{1+n^{-1}}{3n^{-1}})=\infty\). As \(\displaystyle n\) approaches infinity, the negative terms go to zero, which makes zero in the denominator and \(\displaystyle 1\) in the numerator, equalling infinity.)

In order to test the convergence of the series, we use the ratio test.

\(\displaystyle L=\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |\)

If...

\(\displaystyle L< 1\) the series is absolutely convergent

\(\displaystyle L=1\) the convergence is undetermined

\(\displaystyle L>1\) the series diverges

For this problem

\(\displaystyle L=\lim_{n\rightarrow \infty }\left | \frac{\frac{3^{n+1}}{2^{2(n+1)+1}}}{\frac{3^n}{2^{2n+1}}} \right |\)

\(\displaystyle =\lim_{n\rightarrow \infty }\left | \frac{3^{n+1}}{2^{2(n+1)+1} } \cdot \frac{2^{2n+1}}{3^n} \right |\)

\(\displaystyle =\lim_{n\rightarrow \infty }\left | \frac{3\cdot3^{n}}{2^2\cdot2^{2n+1} } \cdot \frac{2^{2n+1}}{3^n} \right |\)

\(\displaystyle =\lim_{n\rightarrow \infty }\left | \frac{3}{2^2} } \right |\)

\(\displaystyle =\frac{3}{4}\)

And since 

\(\displaystyle L< 1\) the series absolutely converges.

an easy sequence to compare \(\displaystyle \frac{1}{n^2+n}\) to is \(\displaystyle \frac{1}{n^2}\), which we know converges.

\(\displaystyle \frac{\frac{1}{n^2}}{\frac{1}{n^2+n}} = \frac{n^2+n}{n^2}\)

We need to find \(\displaystyle \lim_{n \to \infty} \frac{n^2+n}{n^2} = \lim_{n \to \infty} 1+\frac{1}{n}\)

\(\displaystyle \frac{1}{n}\) goes to zero so this limit is \(\displaystyle 1\). Since the limit of the ratio is 1 and \(\displaystyle \sum \frac{1}{n^2}\) converges, \(\displaystyle \sum \frac{1}{n^2+n}\) must converge also.

To use the ratio test, we must first evaluate

\(\displaystyle \lim_{k \to \infty} |\frac{a_{k+1}}{a_k}|\)

\(\displaystyle =\lim_{k \to \infty} |\frac{\frac{(-1)^{k+1}}{(k+1)^3}}{\frac{(-1)^k}{k^3}}|\)

\(\displaystyle =\lim_{k \to \infty} |\frac{(-1)^{k+1}}{(k+1)^{3}} \times \frac{k^3}{(-1)^k}|\)

\(\displaystyle = \lim_{k \to \infty} | -1 \times \frac{k^3}{(k+1)^3}|\)

\(\displaystyle = \lim_{k \to \infty} \frac{k^3}{(k+1)^3}\)

\(\displaystyle =1\)

Since the result of the limit is \(\displaystyle 1\), the Ratio Test fails, and therefore we cannot use it in testing this series for convergence/divergence.

To determine whether the series converges or diverges, we must use the ratio test, which states that for the given series \(\displaystyle \sum_{n=0}^{\infty} a_{n}\), and 

  \(\displaystyle L=\lim_{n\rightarrow \infty} \left |\frac{ a_{n+1}}{a_{n}}\right |\), when L is equal to 1, then the series may be convergent, divergent, or conditionally convergent; when L is less than 1 then the series is convergent; and when L is greater than 1 the series is divergent.

Using the above test, we get

\(\displaystyle L=\lim_{n\rightarrow \infty} \left | \frac{(n+3)!}{5^{n+1}} \cdot \frac{5^n}{(n+2)!} \right |=\lim_{n\rightarrow \infty}\left | \frac{n+3}{5^1}\right |=\infty>1\)

Because L is greater than 1, the series diverges.

The ratio test is defined as follows

\(\displaystyle \lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |\)

where \(\displaystyle a_n\) is the n-th term of the series.

if the limit

• is greater than \(\displaystyle 1\), the series diverges
• is less than \(\displaystyle 1\), the series converges
• equal to \(\displaystyle 1\), the test is inconclusive

Finding the limit for the terms in our series,

\(\displaystyle \lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |=\lim_{n \to \infty} \left | \frac{\frac{(n+1)^2}{3^{n+1}}}{\frac{n^2}{3^n}} \right |\)

\(\displaystyle =\lim_{n \to \infty}\left | \frac{(n^2+2n+1)3^n}{(n^2)3^{n+1}} \right |\)

\(\displaystyle =\frac{1}{3}\lim_{n \to \infty}\left | \frac{n^2+2n+1}{n^2} \right |\)

\(\displaystyle =\frac{1}{3}*1\)

\(\displaystyle =\frac{1}{3}\)

Because the limit is less than \(\displaystyle 1\),

the series converges.

The ratio test is defined as follows

\(\displaystyle \lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |\)

where \(\displaystyle a_n\) is the n-th term of the series.

if the limit

• is greater than \(\displaystyle 1\), the series diverges
• is less than \(\displaystyle 1\), the series converges
• equal to \(\displaystyle 1\), the test is inconclusive

Finding the limit for the terms in our series,

\(\displaystyle \lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |=\lim_{n \to \infty} \left | \frac{\frac{(ln\,n+1)3^{n+1}}{2^{2(n+1)+1}}}{\frac{(ln\,n)3^n}{2^(2n+1)}} \right |\)

\(\displaystyle =\lim_{n \to \infty}\left |\frac{(ln\,n+1)2^{2n+1}*3^{n+1}}{(ln\,n)2^{2n+3}*3^n} \right |\)

\(\displaystyle =\frac{3}{2^2}\lim_{n \to \infty}\left |\frac{ln\,n+1}{ln\,n} \right |\)

\(\displaystyle =\frac{3}{4}*1\)

\(\displaystyle =\frac{3}{4}\)

Because the limit is less than \(\displaystyle 1\),

the series converges.

The ratio test is defined as follows

\(\displaystyle \lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |\)

where \(\displaystyle a_n\) is the n-th term of the series.

if the limit

• is greater than \(\displaystyle 1\), the series diverges
• is less than \(\displaystyle 1\), the series converges
• equal to \(\displaystyle 1\), the test is inconclusive

Finding the limit for the terms in our series,

\(\displaystyle \lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |=\lim_{n \to \infty} \left | \frac{\frac{(n+1)!}{4^{n+1}}}{\frac{n!}{4^n}} \right |\)

\(\displaystyle =\lim_{n \to \infty}\left |\frac{(n+1)n!4^n}{n!4^{n+1}} \right |\)

\(\displaystyle =\frac{1}{4}\lim_{n \to \infty}\left |n+1\right |\)

\(\displaystyle =\frac{1}{4}*\infty\)

\(\displaystyle =\infty\)

Because the limit is greater than \(\displaystyle 1\),

the series diverges.

The ratio test is defined as follows

\(\displaystyle \lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |\)

where \(\displaystyle a_n\) is the n-th term of the series.

if the limit

• is greater than \(\displaystyle 1\), the series diverges
• is less than \(\displaystyle 1\), the series converges
• equal to \(\displaystyle 1\), the test is inconclusive

Finding the limit for the terms in our series,

\(\displaystyle \lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |=\lim_{n \to \infty} \left | \frac{\frac{(n+1)^3}{(n+1)!2^{n+1}}}{\frac{n^3}{n!2^n}} \right |\)

\(\displaystyle =\lim_{n \to \infty}\left | \frac{n!(n+1)^3*2^n}{(n+1)n!n^3*2^{n+1}} \right |\)

\(\displaystyle =\frac{1}{2}\lim_{n \to \infty}\left | \frac{(n+1)^2}{n^3}\right |\)

\(\displaystyle =\frac{1}{2}*0\)

\(\displaystyle =0\)

Because the limit is less than \(\displaystyle 1\),

the series converges.