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
$\displaystyle \frac{7}{(n+1)(n+2)} = \frac{7}{n+1}-\frac{7}{n+2}$
Now,
$\displaystyle \sum_{n=1}^k \frac{7}{n+1}-\frac{7}{n+2} = \left( \frac{7}{2}-\frac{7}{3} \right) + \left( \frac{7}{3}-\frac{7}{4} \right)+....+ \left( \frac{7}{k+1}-\frac{7}{k+2}\right)$
$\displaystyle = \left( \frac{7}{2} -\frac{7}{k+2}\right)$
find the limit as k tends to infinity.
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
$\displaystyle \lim_{n \to \infty}7\left(\frac{1}{2} - \frac{1}{n + 1}\right)$
$\displaystyle = 7\left(\frac{1}{2} - 0\right)$
$\displaystyle = \frac{7}{2}$.