Question

For the following telescoping​ series, find a formula for the nth term of the sequence of partial sums ​{Sn​}. Then evaluate Lim Sn to obtain the value of the series or state that the series diverges.
n→[infinity]
[infinity]
Σ (4/√k+5 ) - 4/ √ k+6)
k=1

Answer 1

Looks like the series is

$\displaystyle\sum_{k=1}^\infty\left(\frac4{\sqrt{k+5}}-\frac4{\sqrt{k+6}}\right)$

This series has n-th partial sum

$S_n=\displaystyle\sum_{k=1}^n\bullet$

(where $\bullet$ is used as a placeholder for the summand)

$S_n=\displaystyle\left(\frac4{\sqrt6}-\frac4{\sqrt7}\right)+\left(\frac4{\sqrt7}-\frac4{\sqrt8}\right)+\cdots+\left(\frac4{\sqrt{n+4}}-\frac4{\sqrt{n+5}}\right)+\left(\frac4{\sqrt{n+5}}-\frac4{\sqrt{n+6}}\right)$

In each grouped term, the last term is annihilated by the first term of the next group; that is, for instance,

$\displaystyle\left(\frac4{\sqrt6}-\frac4{\sqrt7}\right)+\left(\frac4{\sqrt7}-\frac4{\sqrt8}\right)=\frac4{\sqrt6}-\frac4{\sqrt8}$

Ultimately, all the middle terms will vanish and we're left with

$S_n=\dfrac4{\sqrt6}-\dfrac4{\sqrt{n+6}}$

As $n\to\infty$, the last term converges to 0 and we're left with

$\displaystyle\sum_{k=1}^\infty\bullet=\lim_{n\to\infty}S_n=\frac4{\sqrt6}=\boxed{2\sqrt{\dfrac23}}$