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Is $\lim\limits_{k\to\infty}\sum\limits_{n=k+1}^{2k}{\frac{1}{n}} = 0$?

Please help me with this homework question.

Let $\lbrace a_{n} \rbrace$ be the sequence defined by $$a_{n} = \frac{1}{n+1} + \frac{1}{n+2} ... +\frac{1}{2n}$$ for each positive integer $n$. Prove that this sequence converges to ln 2 by showing that $a_{n}$ is related to the partial sums of the series $\displaystyle\sum\limits_{k=0}^\infty (-1)^{k+1}/k.$

I know that $\displaystyle\sum\limits_{k=0}^\infty (-1)^{k+1}/k$ converges to ln 2.

I also know that

$s_{1} = 1,$

$s_{2} = 1/2,$

$s_{3} = 5/6,$

$s_{4} = 7/12,$

etc.

However, I am having a hard time understanding how $\lbrace a_{n} \rbrace$ is related to the partial sum of the given series. Any help or hints are greatly appreciated.

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marked as duplicate by Sivaram Ambikasaran, anon, Byron Schmuland, Jyrki Lahtonen, Zev Chonoles Mar 3 '12 at 2:50

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
You can find multiple (other) proofs in the answers here: math.stackexchange.com/questions/113401/… –  Aryabhata Feb 29 '12 at 23:24

4 Answers 4

Calculate $a_n-a_{n-1}$. It turns out to be $\frac{1}{2n-1}-\frac{1}{2n}$, that is, two successive terms in the power series for $\log 2$.

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$$\begin{align*} 1 - \frac{1}{2} &+ \frac{1}{3} - \dots -\frac{1}{2n} = \\ &=1 + \frac{1}{3} + \frac{1}{5} + \dots + \frac{1}{2n-1} - \left(\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2n}\right) \\ &=1 + \frac{1}{3} + \frac{1}{5} + \dots + \frac{1}{2n-1} - \frac{1}{2}\left(1 + \frac{1}{2} + \dots + \frac{1}{n}\right) \\ &=1 + \frac{1}{3} + \frac{1}{5} + \dots + \frac{1}{2n-1} + \frac{1}{2}\left(1 + \frac{1}{2} + \dots + \frac{1}{n}\right) - \left(1 + \frac{1}{2} + \dots + \frac{1}{n}\right) \\ &=1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{2n} - \left(1 + \frac{1}{2} + \dots + \frac{1}{n}\right) \\ &=\frac{1}{n+1} + \frac{1}{n+2} + \dots +\frac{1}{2n}\phantom{=} \end{align*}$$

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1  
@GreenBeans, clearly this is the answer you are looking for. –  leo Mar 1 '12 at 3:36
    
@leo You spoiled it. GreenBeans, this are the droids you are looking for. –  Pedro Tamaroff Mar 1 '12 at 4:14

This sequence has popped up many times in SE.

You can write it as

$$ s_n = \sum_{k=1}^{n} \frac{1}{n+k}$$

Note that the partial sums of the series for $\log 2$ are

$$\eqalign{ & {p_1} = 1 \cr & {p_2} = 1 - \frac{1}{2} = \frac{1}{2} \cr & {p_3} = \frac{1}{2} + \frac{1}{3} \cr & {p_4} = \frac{1}{2} + \frac{1}{3} - \frac{1}{4} = \frac{1}{3} + \frac{1}{4} \cr & {p_5} = \frac{1}{3} + \frac{1}{4} + \frac{1}{5} \cr & {p_6} = \frac{1}{4} + \frac{1}{5} + \frac{1}{6} \cr & {p_7} = \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} \cr & {p_8} = \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \cr} $$

I guess that with Henning's answer and this, you can probably relate the $s_n$ with $p_n$.


Add: the alternative route is showing that

$$ s_n = \sum_{k=1}^{n} \frac{1}{n+k}$$

$$ s_n = \frac{1}{n}\sum_{k=1}^{n} \frac{1}{1+\frac{k}{n}}$$

Is a Riemman sum for

$$\int_0^1 \frac{dx}{x+1} = \log 2$$

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The answer by Aryabhata is good, but for a more general series you can do the following:

if $s_n = \sum\limits_{k = 1}^{n}f(k)$ , $t_n = \int\limits_{1}^{n}f(x)dx$ , $d_n = s_n - t_n$

where $f$ is a positive decreasing function defined on [1,$+\infty$) such that $\lim_{x\rightarrow\infty}f(x) = 0$ . For n = 1,2,..

then we have

$0\leq d_k - \lim_{n\rightarrow \infty}d_n \leq f(k)$ for k = 1,2,...

You can check this directly, or you can refer Mathematical Analysis Apostol pg. 191 Thm 8.23

Let D = $ \lim_{n\rightarrow \infty}d_n$

So, $0 \leq s_n - t_n - D \leq f(n)$

Or we can write $s_n = t_n + D + O(f(n))$

Hence directly you get for $f(x) = \frac{1}{x}$

$\sum\limits_{k = 1}^{n} \frac{1}{k} = \ln(n) + D + O(\frac{1}{n})$

Using this formula you can show that for $p,q$ intergers s.t. $p \geq q\geq 1$

$\lim_{n\rightarrow \infty}\sum\limits_{k = qn +1}^{pn} \frac{1}{k} = \ln(\frac{p}{q})$

for q = 1, p = 2 you get $\ln 2$

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