Find a formula for the nth partial sum of the series and use it to find the seriesʹ sum if the series converges.

1)

I have no idea how to do this one. I missed it on the test and it might show up on the final

Surely this is

\(\displaystyle 7\sum_{n = 2}^{\infty}\frac{1}{n(n + 1)}\).

Using partial fractions:

\(\displaystyle \frac{A}{n} + \frac{B}{n + 1} = \frac{1}{n(n + 1)}\)

\(\displaystyle \frac{A(n + 1) + Bn}{n(n + 1)} = \frac{1}{n(n + 1)}\)

\(\displaystyle A(n + 1) + Bn = 1\)

\(\displaystyle (A + B)n + A= 0n + 1\).

So \(\displaystyle A = 1\) and \(\displaystyle B = -1\).

Therefore we can write the sum as

\(\displaystyle 7\sum_{n = 2}^{\infty}\frac{1}{n(n + 1)} = 7\sum_{n = 2}^{\infty}\left(\frac{1}{n} - \frac{1}{n + 1}\right)\)

This is a telescoping series, so all terms except the first and last will cancel.

So the \(\displaystyle n^{\textrm{th}}\) partial sum is

\(\displaystyle 7\left(\frac{1}{2} - \frac{1}{n + 1}\right)\).

Therefore, the sum of the series is

\(\displaystyle \lim_{n \to \infty}7\left(\frac{1}{2} - \frac{1}{n + 1}\right)\)

\(\displaystyle = 7\left(\frac{1}{2} - 0\right)\)

\(\displaystyle = \frac{7}{2}\).