# Prove $\sum_{k=1}^n \frac{1}{(2k-1)(2k+1)}=\frac{n}{2n+1}$ [closed]

I have attached an image of a kind of mathematical induction question that i have never seen before. I attached it because i don't know how to type all the symbols out properly, i'm sorry again would really appreciate any help :)

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

Use this to evaluate $$\sum_{k=13}^{37} \frac{1}{4k^2-1}$$

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• What was your attempt? – Mayank Deora Jun 1 '16 at 9:32
• You have to induction for n: This hold for $n=1$. Assume this hold for $n=x$, then prove it holds for $n=x+1$. Thus it hold for all $n\in \mathbb N$. – user202729 Jun 1 '16 at 9:33
• Once you do the induction, note that $$\sum_{j=13}^{37}=\sum_{j=1}^{37}-\sum_{j=1}^{12}$$ – Eleven-Eleven Jun 1 '16 at 9:33
• Have you proved the identity by induction? Actually i don't understand what are you saying in the above comment , please elaborate. – Mayank Deora Jun 1 '16 at 9:44
• I don't understand this problem well enough to be able to solve it – Daniel Jun 1 '16 at 9:48

## 5 Answers

A full proof using induction:

1. Check if the formula holds for $n=1$.

$$\sum_{k=1}^1 \frac{1}{(2k-1)(2k+1)}=\frac{1}{(2\cdot 1-1)(2\cdot 1+1)}=\frac{1}{3}$$ $$\frac{1}{2\cdot 1+1}=\frac{1}{3}$$ So the formula holds for $n=1$.

1. Assume the formula holds for $n$. We want to prove that it holds for $n+1$. So assume that $$\sum_{k=1}^{n} \frac{1}{(2k-1)(2k+1)}=\frac{n}{2n+1}$$ and we want to prove $$\sum_{k=1}^{n+1} \frac{1}{(2k-1)(2k+1)}=\frac{n+1}{2(n+1)+1}$$

Observe that $$\sum_{k=1}^{n+1} \frac{1}{(2k-1)(2k+1)}=\sum_{k=1}^{n} \frac{1}{(2k-1)(2k+1)}+\frac{1}{(2(n+1)-1)(2(n+1)+1)}=\dots$$ (the last term taken out of the sum)

and now, by our assumption we have $$\dots = \frac{n}{2n+1}+\frac{1}{(2(n+1)-1)(2(n+1)+1)}$$

To finish the proof, you only need to check the identity @user243301 wrote.

• thank you so much that helps alot :)!! – Daniel Jun 1 '16 at 14:02

Let me address your problem one-by one:

• The "sigma" denoted by $\sum$ actually means summation, if you already did not know it. Using this sign or operator as you wish to call it, we can rewrite $a_1+a_2+a_3+\ldots +a_n$ as$\sum_{k=1}^n a_k$.

• The induction comes up easily:

In this case, the mathematical statement $$P(n) : \sum_{k=1}^n \frac{1}{(2k-1)(2k+1)}=\frac{n}{2n+1}$$

First of all, as given, we have that L.H.S.$=\sum_{k=1}^1 \frac{1}{(2k-1)(2k+1)}=\frac{1}{(2-1)(2+1)}=\frac{1}{2+1}=$R.H.S.
So $P(1)$ is true.

Now, using induction, assume that $P(m)$ is true i.e. $$\sum_{k=1}^m \frac{1}{(2k-1)(2k+1)}=\frac{m}{2m+1}$$

Thus, we can say that \begin{align} & \sum_{k=1}^{m+1} \frac{1}{(2k-1)(2k+1)} \\ & =\sum_{k=1}^m \frac{1}{(2k-1)(2k+1)} + \frac{1}{[2(m+1)-1][2(m+1)+1]} \\ & =\frac{m}{2m+1} + \frac{1}{(2m+1)(2m+3)} \\ & =\frac{m(2m+3)+1}{(2m+1)(2m+3)} \\ & =\frac{2m^2+3m+1}{(2m+1)(2m+3)} \\ & =\frac{(2m+1)(m+1)}{(2m+1)(2m+3)} \\ & =\frac{(m+1)}{2(m+1)+1}\end{align}

So $P(m+1)$ is true.

Hence, the truth of $P(m) \Rightarrow P(m+1)$ is true and the statement given is proved to be true by principle of mathematical induction.

• The second part of your answer comes up from the above result:

\begin{align} & \sum_{k=13}^{37} \frac{1}{4k^2-1} \\ & =\sum_{k=1}^{37} \frac{1}{4k^2-1}-\sum_{k=1}^{12} \frac{1}{4k^2-1} \\ & =\sum_{k=1}^{37} \frac{1}{(2k-1)(2k+1)}-\sum_{k=1}^{12} \frac{1}{(2k-1)(2k+1)} \\ & =\frac{37}{74+1}-\frac{12}{24+1}\end{align}

I hope you now understand.

• i dont fullu understand but thats not your fault thank you so much for your detailed explanation – Daniel Jun 1 '16 at 10:02
• @Daniel Please elaborate where or in which steps or statements you are having trouble in understanding my solution. – SchrodingersCat Jun 1 '16 at 10:04
• im confused about where the 2m+3 comes from. For someone who is very new to these types of questions what would be good practice for me? – Daniel Jun 1 '16 at 10:07
• @Daniel First of all, there is only one particular way of proceeding in these kinds of induction problems and it is same for all, expert or amateur or young or old. I have also followed that same procedure. And secondly, in the given expression,$\frac{1}{(2k-1)(2k+1)}$, what happens if you put $k=m+1$? That is just what I have done. – SchrodingersCat Jun 1 '16 at 10:13

Hints: As your know first you need check the first identity for $n=1$. After presume that for $n$ the identity holds. Then you need to show $n\implies n+1$, write the details that why we need to show $$\frac{n}{2n+1}+\frac{1}{(2(n+1)-1)(2(n+1)+1)}=\frac{(n+1)}{2(n+1)+1}.$$ You need to divide polynomials or relate $(n+1)(2n+1)=2n^2+3n+1$ with previous computation. The second question is an specialization of the first identity, since we know the remarkable identity involving the quotients. Welcome to this site, also try next time show your effort and write in TEX, is easy with this Tutorial

• @Daniel There was a typo, that was fixed. Daniel you know do these computations: write it, is only type the formula for $n$ and after do a comparison with the formula for $n+1$, this is the specialization of formula with the upper index $n+1$. Split this last sum to can, to use that you know the sum first sum, it is $n/(2n+1)$. I can not say more, because my english is bad I've put so many hints in your hands (read examples in your books to do a comparison because these examples have in common a structure). – user243301 Jun 1 '16 at 9:53
• my books dont have induction questions like this they are much simplier but thank you for your help – Daniel Jun 1 '16 at 9:55
• You need to see severals simple proofs, in this site you should be find their. I am sorry by my tone, but the only manner that you understand it is with several detailed examples, and takes away your hand over your eyes to see that it is easy. I hope that other user provide to you a more detailed answer. Good look @Daniel – user243301 Jun 1 '16 at 9:59
• A vote up @SchrodingersCat – user243301 Jun 1 '16 at 10:01

This is a solution without induction, if you're interested in another method working in such cases (to do induction, you have to know the result, while telescoping sum works without it).

Observe that $$\frac{1}{(2k-1)(2k+1)}=\frac{1}{2}\left(\frac{1}{2k-1}-\frac{1}{2k+1}\right)$$ Hence your sum telescopes:

$$\sum_{k=1}^{n}\frac{1}{(2k-1)(2k+1)}=\frac{1}{2}\left(\frac{1}{2\cdot 1-1}-\frac{1}{2\cdot n+1}\right)=\frac{n}{2n+1}$$ (only first and last term are not cancelled)

see some examples of telescoping sums, for example

https://en.wikipedia.org/wiki/Telescoping_series

Help with telescoping sum

• Thank you @user340508, do i still follow the normal steps for n = base case then n= k and finally n = k+1, the confusion is i dont know how to implement them. I know it feels like you are making it clear but i reall do not understand – Daniel Jun 1 '16 at 9:52
• @Daniel $$(1-\color{blue}{\frac12})+(\color{blue}{\frac12}-\color {red}{\frac13})+(\color {red}{\frac13}-\frac14)+\dots$$ Basic telescoping series. Terms cancel. – Simply Beautiful Art Jun 1 '16 at 11:38
• And I say that telescoping series is just a neat way to not say induction, but nicely spotted – WorldSEnder Jun 1 '16 at 13:23
• @Daniel Telescopy is a form of induction specialized to such sums. I show how to formalize it inductively in my answer here. – Bill Dubuque Jun 1 '16 at 14:04
• @ Bill Dubuque , yes i saw thank you i think i just need to practice this type of induction abit more – Daniel Jun 1 '16 at 14:06

Hint $\$ The sum telescopes, i.e. the induction step simplifies as follows

If $\qquad f(1) + f(2)+\cdots + f(n\!-\!1)\quad =\quad \color{#0a0}{g(n\!-\!1)}$

then $\ \ \ f(1) + f(2)+\cdots +f(n\!-\!1) + \color{#c00}{f(n)}\,=\,\color{#0a0}{ g(n)}\iff \color{#c00}{f(n)}=\color{#0a0}{g(n)-g(n\!-\!1)}$

Combining the above with the base case $\,f(1)=g(1)\,$ we obtain an inductive proof of

$$\sum_{k=1}^{n} f(k)\, =\, g(n)\,\iff\, \underbrace{f(n) = g(n)-g(n\!-\!1)}_{\large \rm inductive\ step}\ {\rm and}\ \underbrace{f(1) = g(1)}_{\large\rm base\ case}$$

In your case you have $\,g(n) = n/(2n\!+\!1)\,$ so, by above, the induction step holds for this $\,g(n)\,$ iff $\, g(n)-g(n\!-\!1) = f(n) = 1/(2n\!-\!1)(2n\!+\!1),\,$ which is easily verified .

Remark $\$ My prior posts contain many more examples of telescopy. It reduces problems like this to trivial high-school arithmetic, i.e. verifying the equality of two polynomials in $\,n$.