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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

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up vote 4 down vote accepted

The sum is less than $\sum_{n=2}^{\infty} \frac{1}{2^n} = \frac{1}{2}$.

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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.


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

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+1 for the spoiler! I hadn't saw this before. – 000 Dec 29 '12 at 0:51
*hadn't seen =P – Pedro Tamaroff Dec 29 '12 at 1:15

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.

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"positive" $\mapsto$ "$\ge 1$". Also, you are describing finite sums, so there is never any question of convergence (boundedness, sure). – Erick Wong Dec 29 '12 at 1:42
Nite approach +1 – Babak S. Dec 29 '12 at 7:16

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:

$$m \leq 2^{m+1}$$

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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.

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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 $\ln(1-x), |x|<1)$


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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.

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How do you know that the last inequality is true? – Jmaff Dec 29 '12 at 2:34
Is induction required on n and m to show that? – Jmaff Dec 29 '12 at 2:48
No, only the observation I make in my answer. – Pedro Tamaroff Dec 29 '12 at 5:02
sorry to get back to this late, but i mean the portion" $\sum_{k=m+1}^{n} \frac {1}{2^{k}} < \frac{1}{2^{m}}$" – Jmaff Dec 29 '12 at 21:23

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 $\ln{2}$.

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