\(\displaystyle \sum_{n= 1}^{\infty } \left ( \frac{e-1}{e} \right )^n\) is an infinite geometric series with initial term \(\displaystyle \left ( \frac{e-1}{e} \right )^1 = \frac{e-1}{e}\) and common ratio \(\displaystyle \frac{e-1}{e}\). The sum is therefore

\(\displaystyle \sum_{n= 1}^{\infty } \left ( \frac{e-1}{e} \right )^n\)

\(\displaystyle =\frac{ \frac{e-1}{e} }{1-\frac{e-1}{e}}\) \(\displaystyle \frac{}{}\)

\(\displaystyle =\frac{ e \cdot \frac{e-1}{e} }{e \cdot \left ( 1-\frac{e-1}{e} \right )}\)

\(\displaystyle = \frac{e-1}{e-(e-1)} = \frac{e-1}{1} = e - 1\)

\(\displaystyle \sum_{n=0}^{\infty } \left (-\frac{1}{e} \right )^n\) is a geometric series with initial term \(\displaystyle \left (-\frac{1}{e} \right )^0 = 1\) and common ratio \(\displaystyle -\frac{1}{e}\). The sum of this series is 

\(\displaystyle \frac{1}{1-\left ( -\frac{1}{e} \right )}= \frac{1}{1+ \frac{1}{e} \right )} = \frac{e}{e+1\right )}\).

In order to uniquely define the geometric series, we need to know two things: the ratio between successive terms and at least one of the terms. Knowing one term doesn't give you the ratio of successive terms, but knowing two terms will give you the ratio. By the term generator for a geometric series \(\displaystyle \small a_{n}=a_{1}r^{n-1}\), you can see that you only need two terms to find the ratio \(\displaystyle \small r\). 

The sum formula for \(\displaystyle \small n\) terms of an arithmetic series is \(\displaystyle \small (a_{1}+a_{n})(n/2)\).

For \(\displaystyle \small 20\) terms, this formula becomes \(\displaystyle \small \small \small (a_{1}+a_{20})(20/2)\).

Using our term generator for \(\displaystyle \small a_{1}\) and \(\displaystyle \small a_{20}\), this formula becomes 

\(\displaystyle \small \small \small \small \small (3(1)-2+3(20)-2)(20/2)=(59)(10)=590\).

We can evaluate the infinite series by recognizing it as a geometric series times some constant.

Let's manipulate this series:

\(\displaystyle \small \small \small \sum_{j=2}^\infty (-1)^j\left(\frac{1}{2}\right)^{j+1}=\sum_{j=2}^\infty (-1)^2(-1)^{j-2}\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^{j-2}\)

\(\displaystyle \small \small \small =(-1)^2\left(\frac{1}{2}\right)^3\sum_{j=2}^\infty (-1)^{j-2}\left(\frac{1}{2}\right)^{j-2}=\frac{1}{8}\sum_{j=0}^\infty \left(\frac{-1}{2}\right)^{j}\).

Now it suffices to evaluate \(\displaystyle \small \small \small \sum_{j=0}^\infty \left(\frac{-1}{2}\right)^{j}\), which we can recognize as the power series of \(\displaystyle \small \frac{1}{1-x}\) with \(\displaystyle \small x=-\frac{1}{2}\), which is

 \(\displaystyle \small \small \small \sum_{j=0}^\infty \left(\frac{-1}{2}\right)^{j}=\left( \frac{1}{1-(-1/2)}\right)=\left(\frac{1}{1.5} \right )=\frac{2}{3}\).

So we have

\(\displaystyle \small \small \sum_{j=2}^\infty (-1)^j\left(\frac{1}{2}\right)^{j+1}=\frac{1}{8}\cdot \frac{2}{3}=\frac{1}{12}\).

If a series is arithmetic, then there exists a common difference between each pair of consecutive terms in the series.

For this series

\(\displaystyle a_1=1,\, a_2=8,\,a_3=27,\,a_4=64\)

Because 

\(\displaystyle a_2-a_1=8-1=7\)

and

\(\displaystyle a_3-a_2=27-8=19\)

we find that there does NOT exist a common difference and as such,

the series is not arithmetic.

To determine the convergency of a geometric series, we must find the absolute value of the common ratio.

A geometric series will converge if the absolute value of the common ratio is less than one, or

\(\displaystyle |r|< 1\)

In this problem, we see that 

\(\displaystyle r=\pi \approx 3.14\)

And because \(\displaystyle |\pi|>1\)

we conclude that the series does not converge to a finite sum.

This is an arithmetic series.

Its general form is \(\displaystyle \sum^{n}_{k=1}a_k\).

To calculate the sum of very large series such as these, use the formula \(\displaystyle \frac{n}{2}(a_1 + a_n)\)

This works because you are taking the average of the largest and smallest terms, and then multiplying them by n, which is the same as calculating the sum total.

Solution:

\(\displaystyle a_{1} = 2*1 + 3 = 5\)

\(\displaystyle a_{500} = 2 * 500 +3 = 1000 + 3 = 1003\)

\(\displaystyle n =500\)

\(\displaystyle \sum^{500}_{k=1}2k+3= \frac{n}{2}(a_{1}+a_{500}) = \frac{500}{2}(5+1003)=250*1008=252000\)

This is an arithmetic series.

Its general form is \(\displaystyle \sum^{n}_{k=1}a_k\)

To calculate the sum of very large series such as these, use the formula \(\displaystyle \frac{n}{2}(a_1 + a_n)\)

This works because you are taking the average of the largest and smallest terms, and then multiplying them by n, which is the same as calculating the sum total.

Solution:

\(\displaystyle a_{1} = 3*1+4 = 3+4=7\)

\(\displaystyle a_{300} = 3*300+4 = 900+4 = 904\)

\(\displaystyle n = 300\)

\(\displaystyle \sum^{300}_{k=1}3k+4= \frac{n}{2}(a_{1}+a_{300}) = \frac{300}{2}(7+904)=150*911=136650\)

This is an arithmetic series.

Its general form is \(\displaystyle \sum^{n}_{k=1}a_k\)

To calculate the sum of very large series such as these, use the formula \(\displaystyle \frac{n}{2}(a_1 + a_n)\)

This works because you are taking the average of the largest and smallest terms, and then multiplying them by n, which is the same as calculating the sum total.

Solution:

\(\displaystyle a_{1}=6*1=6\)

\(\displaystyle a_{1200}=6*1200 = 7200\)

\(\displaystyle n =1200\)

\(\displaystyle \sum^{1200}_{k=1}6k= \frac{n}{2}(a_{1}+a_{1200}) = \frac{1200}{2}(6+7200)=600*7206 = 4323600\)