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There are both $n$ and $k$ in this formula and I don’t know how to prove it. Because in this case I cannot split the summation into parts contains the assumption.

$$\sum_{k=1}^n \frac{1}{(2k-1)2k} = \sum_{k=1}^n \frac{1}{n+k}$$

My basic case is $n=1$ and both LHS and RHS equals to $\frac{1}{2}$. Assume the statement is true for some m $$ie. \sum_{k=1}^m \frac{1}{(2k-1)2k} = \sum_{k=1}^m \frac{1}{m+k}$$ I want to prove for $m+1$ $$ie. \sum_{k=1}^{m+1} \frac{1}{(2k-1)2k} = \sum_{k=1}^{m+1} \frac{1}{m+1+k}$$ RHS I got $$\sum_{k=1}^{m+1} \frac{1}{m+1+k}=\sum_{k=1}^{m} \frac{1}{m+1+k}+\frac{1}{m+1+m+1}=\sum_{k=1}^{m} \frac{1}{m+1+k}+\frac{1}{2m+2}$$ LHS i got $$\sum_{k=1}^{m+1} \frac{1}{(2k-1)2k} = \sum_{k=1}^{m} \frac{1}{(2k-1)2k} + \frac{1}{(2(m+1)-1)2(m+1)}=\sum_{k=1}^m \frac{1}{m+k}+\frac{1}{(2m+2)(2m+1)}$$ And I got stuck here.Because RHS it is $m+1+k$ on the denominator but on LHS it is $m+k$ and they are quite different I don't know how to prove them equal.

SixTwelfthsPi
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1 Answers1

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You actually can split it.

HINT: $$\frac1{2k(2k-1)}=\frac1{2k-1}-\frac1{2k}\,,$$

so

$$\sum_{k=1}^{n+1}\frac1{2k(2k-1)}-\sum_{k=1}^n\frac1{2k(2k-1)}=\frac1{2n+1}-\frac1{2(n+1)}\,.$$

Edit: And

$$\begin{align*} \sum_{k=1}^{n+1}\frac1{n+1+k}-\sum_{k=1}^n\frac1{n+k}&=\sum_{k=2}^{n+2}\frac1{n+k}-\sum_{k=1}^n\frac1{n+k}\\ &=\frac1{2n+1}+\frac1{2n+2}-\frac1{n+1}\,. \end{align*}$$

The only terms in the first summation that do not appear in the second are the last two, with $k=n+1$ and $k=n+2$, and the only term in the second summation that does not appear in the first is the $k=1$ term: everything else cancels out. Thus, you’ll have the induction step once you show that

$$\frac1{2n+1}-\frac1{2(n+1)}=\frac1{2n+1}+\frac1{2n+2}-\frac1{n+1}\,,$$

which is just a matter of a little algebra.

Brian M. Scott
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  • $$\sum_{k=1}^n \frac{-1}{(n+1+k)(n+k)}+\frac{1}{2n+2}$$ this is what I got when I tried to figure out the second equation. Still I found noway out. Sorry I am not so clever. Could you please be more specific? – SixTwelfthsPi Mar 06 '21 at 22:01
  • @lllll: I’d prefer not to tell you directly right away. First I’d like you to try writing out the case $n=2$, expanding the summations in full; that should make it a little easier to see what’s going on. For the difference that I asked you to figure out you should get three terms, two positive and one negative. – Brian M. Scott Mar 06 '21 at 22:17
  • Yes I wrote out the case $n=2$ they both equal to $\frac{7}{12}$. And the difference is it $\sum_{k=1}^{n} \frac{1}{(n+k+1} + \frac{1}{2(n+1)} - \sum_{k=1}^{n} \frac{1}{n+k}$? I need to prove the two differences equal, right? But how should I get rid of the summation in the second difference? – SixTwelfthsPi Mar 06 '21 at 22:45
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    @lllll: I see now: you’re going about it wrong. I showed you how much the lefthand sum increases when you go from $n$ to $n+1$; my question was designed to get you to calculate how much the righthand sum increases when you go from $n$ to $n+1$. If you do that correctly, a very little algebra is needed to show that the increases are the same. That gives you your induction step: if the sums are equal for $n$, and they increase by the same amount when you change $n$ to $n+1$, then they’re equal for $n+1$ as well. Does that help? – Brian M. Scott Mar 06 '21 at 22:59
  • I can understand the increase by the same part. But I really don't know how to calculate out the correct difference.;((( – SixTwelfthsPi Mar 06 '21 at 23:19
  • @lllll: Okay; give me a few minutes, and I’ll add that calculation. – Brian M. Scott Mar 06 '21 at 23:20
  • Okay. Thank you very much. – SixTwelfthsPi Mar 06 '21 at 23:23
  • @lllll: Done: I’ve left just a little bit for you at the end. – Brian M. Scott Mar 06 '21 at 23:28
  • Yes!! I see how to do it now! Thank you!! – SixTwelfthsPi Mar 06 '21 at 23:34
  • @lllll: You’re welcome! – Brian M. Scott Mar 06 '21 at 23:34