# How do you use the limit comparison test to determine if Sigma n/((n+1)2^(n-1)) from [1,oo) is convergent or divergent?

Nov 20, 2016

${\sum}_{n = 1}^{\infty} \frac{n}{\left(n + 1\right) {2}^{n - 1}}$

Rewriting this:

$= {\sum}_{n = 1}^{\infty} \frac{n}{n + 1} \frac{1}{{2}^{n - 1}} = {\sum}_{n = 1}^{\infty} \frac{n}{n + 1} \frac{1}{{2}^{n} / 2} = {\sum}_{n = 1}^{\infty} \frac{n}{n + 1} \frac{2}{2} ^ n$

Bringing the $2$ out and noticing that $\frac{1}{2} ^ n = {\left(\frac{1}{2}\right)}^{n}$:

$= 2 {\sum}_{n = 1}^{\infty} \frac{n}{n + 1} {\left(\frac{1}{2}\right)}^{n}$

We should recognize that ${\sum}_{n = 1}^{\infty} {\left(\frac{1}{2}\right)}^{n}$ is a geometric series, and since $\frac{1}{2} < 1$, we know this series will converge.

Also note that when $n > 0$, all terms of $\frac{n}{n + 1} < 1$. This means we can say that:

$\frac{n}{n + 1} {\left(\frac{1}{2}\right)}^{n} \le {\left(\frac{1}{2}\right)}^{n}$

We can now use the direct comparison test. Since ${\sum}_{n = 1}^{\infty} {\left(\frac{1}{2}\right)}^{n}$ converges, and $\frac{n}{n + 1} {\left(\frac{1}{2}\right)}^{n} \le {\left(\frac{1}{2}\right)}^{n}$, we know that ${\sum}_{n = 1}^{\infty} \frac{n}{n + 1} {\left(\frac{1}{2}\right)}^{n}$ converges as well.

Thus $2 {\sum}_{n = 1}^{\infty} \frac{n}{n + 1} {\left(\frac{1}{2}\right)}^{n} = {\sum}_{n = 1}^{\infty} \frac{n}{\left(n + 1\right) {2}^{n - 1}}$ is convergent.