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Please help me to prove that $\gamma_{n} = 1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{n}-\ln n$ converge and it's limit $\gamma \in [0,1]$. Then, using $\gamma_{n}$ find the sum: $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n}$.

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Welcome to MSE! I'd suggest though you'd ask question in a less imperative way and in a more friendly way. Plus, have you managed to find something? Any approaches? What I mean is, you're not a book, you're human, so act like one, it makes us more willing to answer. =) –  Patrick Da Silva Aug 10 '11 at 11:49
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The following question addresses the convergence –  Sasha Aug 10 '11 at 11:56
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possible duplicate of What is $\lim_{n \to \infty}\sum_{k=1}^n 1/k$ –  user17762 Aug 10 '11 at 12:14
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Sorry because my question was imperative. My english is not very good and I translated word by word from my language. I will be more carefully next time. –  NumLock Aug 10 '11 at 12:28
    
In the sense that if you read the answer given, all the good stuff's in there. Although they don't compute the alternated sum. I don't agree it's a duplicate. –  Patrick Da Silva Aug 10 '11 at 12:56

4 Answers 4

up vote 5 down vote accepted

The following question addresses the convergence, so assume $\gamma_n$ converges and denote the limit as $\gamma$.

Now consider $\sum_{n=1}^{2m} (-1)^{n+1} \frac{1}{n}$ and split it into even and odd terms $\sum_{n=1}^m \frac{1}{2n-1} - \sum_{n=1}^m \frac{1}{2n}$. Then complete the sum over odd integers to sum over consequtive integers: $\sum_{n=1}^{2m} \frac{1}{n} - 2 \sum_{n=1}^m \frac{1}{2n}$.

Then subtract logarithms to form $\gamma_{2m} - 2 \gamma_m + \log(2)$, like so $$ ( \sum_{n=1}^{2m} \frac{1}{n} - \log(2m)) - ( \sum_{n=1}^m \frac{1}{n} - \log(m)) + \log(2).$$

In the limit $m \to \infty$ it becomes $ \gamma - \gamma + \log(2) = \log(2)$.

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Thank you for your help too! –  NumLock Aug 10 '11 at 12:34

You can use the mean value theorem to prove first that $$\frac{1}{n+1}<\ln(n+1)-\ln n<\frac{1}{n}$$ then, $\ln(n+1)<1+\ldots+\frac{1}{n}<1+\ln n$, and finally $$\ln(n+1)-\ln n<\gamma_n<1$$ which implies the convergence of $\gamma_n$ and its limit $\gamma\in[0,1]$.

For the second part, I am not sure if you can use $\gamma_n$ to find the sum $\sum_{n=1}^\infty(-1)^{n+1}\frac{1}{n}$!

I suggest to use the Taylor-Young formula for $x\mapsto\ln(x+1)$ on $[0,1]$ to prove that $$\sum_{n=1}^\infty(-1)^{n+1}\frac{1}{n}=\ln2.$$

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Thank you !I learnt mean value theorem in highschool but I completely forgot about it. –  NumLock Aug 10 '11 at 12:34
    
@amine it's mercator series time... –  user38268 Aug 10 '11 at 12:58

The Mean Value Theorem says $$ \frac{1}{k+1}<\log(k+1)-\log(k)<\frac{1}{k} $$ Summing from $m$ to $n-1$ ($m < n$), we get $$ \begin{align} \sum_{k=m}^{n-1}\frac{1}{k+1}<\log(n)-\log(m)<\sum_{k=m}^{n-1}\frac{1}{k} \end{align} $$ Subtracting $\displaystyle{\sum_{k=m}^{n-1}\frac{1}{k+1}=\sum_{k=m+1}^n\frac{1}{k}=\sum_{k=1}^n\frac{1}{k}-\sum_{k=1}^m\frac{1}{k}}$ from all parts, we get $$ 0<\left(\log(n)-\sum_{k=1}^n\frac{1}{k}\right)-\left(\log(m)-\sum_{k=1}^m\frac{1}{k}\right)<\frac{1}{m}-\frac{1}{n} $$ This proves the existence of $\displaystyle{\lim_{n\to\infty}\left(\log(n)-\sum_{k=1}^n\frac{1}{k}\right)}$. If we set $m=1$ and subtract all sides from $1$, we get $$ 1>\left(\sum_{k=1}^n\frac{1}{k}-\log(n)\right)>\frac{1}{n} $$Thus, we have shown that $\displaystyle{\gamma=\lim_{n\to\infty}\left(\sum_{k=1}^n\frac{1}{k}-\log(n)\right)}$ exists, and that $0\le\gamma\le 1$.

Note that $$ \begin{align} \sum_{k=1}^{2n}(-1)^{k+1}\frac{1}{k}&=\sum_{k=1}^{2n}\frac{1}{k}-2\sum_{k=1}^n\frac{1}{2k}\\ &=\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^n\frac{1}{k}\\ &=\left(\sum_{k=1}^{2n}\frac{1}{k}-\log(2n)\right)-\left(\sum_{k=1}^n\frac{1}{k}-\log(n)\right)+\log(2) \end{align} $$ Taking the limit as $n\to\infty$, we get $$ \sum_{k=1}^\infty(-1)^{k+1}\frac{1}{k}=\gamma-\gamma+\log(2)=\log(2) $$

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I apologize if this is close to Sasha's answer, but I have been working on this, during breaks, for quite a while, and posted it before I noticed. Since I show more of the details, I will leave it for now. I can remove it if requested. –  robjohn Aug 10 '11 at 17:26
    
Sorry, for the convergence, I forgot to say that $\gamma_n$ is decreasing because $$\gamma_{n+1}-\gamma_n=\frac{1}{n+1}-(\ln(n+1)-\ln n)<0$$ since it is bounded, it is convergent! –  amine Aug 10 '11 at 21:49

Do we have a "proof without words" question here? I don't find it. There is a lovely diagram found in certain calculus texts that does this for us. Maybe I will animate it, for fun.

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I am not getting this. Do you mean pictorial proof, which suggests the proof ? –  Sasha Aug 10 '11 at 14:15
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It's something like the diagram at the top right here: en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant but then you slide all the blue regions to the left to see that they all fit inside a $1 \times 1$ square. –  GEdgar Aug 10 '11 at 14:54

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