# nth partial sum

#### krzyrice

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

#### chisigma

MHF Hall of Honor
The 'key' is the identity...

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

Kind regards

$$\displaystyle \chi$$ $$\displaystyle \sigma$$

#### krzyrice

i'm still not getting it

#### harish21

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.

• krzyrice

#### krzyrice

thanks makes sense now

#### Prove It

MHF Helper
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

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

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

$$\displaystyle = \frac{7}{2}$$.

• krzyrice