Want to show $\sum_{n=2}^{\infty} \frac{1}{2^{n}*n}$ converges Want to show $\sum_{n=2}^{\infty} \frac{1}{2^{n}*n}$ converges. I am trying to show this by showing that the partial sums are bounded. I have tried doing this by induction but am not seeing how to pass the inductive assumption part. Do I need to instead look for a closed form? thanks
 A: Hint All the terms are postive and $$\frac{1}{n2^n}\leq \frac 1 {2^n}$$ 
Think about the monotone convergence, or the so called comparison test.
Spoiler

 The partial sums are bounded above, and are monotone increasing, thus the monotone convergence theorem implies convergence.

A: You might want to realize a general principle here.
If we have $$\sum_{a \le n \le b}\frac{1}{f(n)g(n)}$$ and $f(n)$ and $g(n)$ are positive on $[a,b]$, then we always have that $$\sum_{a \le n \le b}\frac{1}{f(n)g(n)}\le\sum_{a\le n\le b}\frac{1}{f(n)} \text{ and } \sum_{a \le n \le b}\frac{1}{f(n)g(n)}\le\sum_{a \le n \le b}\frac{1}{g(n)}$$ when the sums $$\sum_{a \le n \le b}\frac{1}{f(n)}$$ and $$\sum_{a \le n \le b}\frac{1}{g(n)}$$ both converge.
In our case, we have that only one converges. Thus, only one of the inequalities is true.
A: The sum is less than $\sum_{n=2}^{\infty} \frac{1}{2^n} = \frac{1}{2}$.
A: Hint: 
Try using the root test. ${}{}{}$ This will give you that the series converges.
You need only know that $\lim_{n\to \infty}\sqrt[\large n]{n} = 1.\quad$ Then
$$\lim_{n\to \infty} \sqrt[\large n]{\dfrac{1}{n2^n}} = \sqrt[\large n]{\frac{1}{n}}\cdot\sqrt[\large n]{\frac{1}{2^n}} = 1 \cdot \dfrac{1}{2} = \dfrac{1}{2} < 1 \implies \sum_{n=2}^{\infty} \frac{1}{n\cdot2^{n}}\quad \text{converges}.$$
Note also that you'd get the same result (that the series converges), by using the the ratio test.  Both the root test and ratio test are very useful for establishing convergence and/or divergence of many series.
A: Since you mentioned induction:
Let $s_m = \sum_{n=2}^{m} \frac{1}{2^{n}*n}$. Then $s_m \leq 1-\frac{1}{m}$.
$P(2)$ is obvious, while $P(m) \Rightarrow P(m+1)$ reduces to
$$1-\frac{1}{m}+\frac{1}{2^{m+1}(m+1)} \leq 1- \frac{1}{m+1}$$
which is equivalent to:
$$m \leq 2^{m+1}$$
A: Or we can go this way
$$\sum_{n=2}^{\infty} \frac{1}{n2^{n}}\le\sum_{n=2}^{\infty} \frac{1}{n^2}\le \sum_{n=2}^{\infty} \frac{1}{n(n-1)}=1$$
and the sum is precisely $\log 2-\frac{1}{2}$ (hint: use the Taylor expansion of $\ln(1-x), |x|<1)$
Chris.
A: Go back to the Cauchy test for convergence of a sequence.
Let $S_n = \sum_{k=2}^{n} \frac{1}{k2^k}$.
If $m < n$,
$S_n-S_m = \sum_{k=m+1}^{n} \frac{1}{k2^k}
< \sum_{k=m+1}^{n} \frac{1}{2^k}
< \frac{1}{2^m}
$
and this $\to 0$
as $m \to \infty$.
Therefore the sequence converges.
A: You can show that $-\ln(1-x)=\sum_1^\infty {x^n \over n}$ for $-1\le x<1$. Hence the series in question converges to $\ln{2}$.
