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

Question

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 $$\sum_{k=13}^{37} \frac{1}{4k^2-1}$$

Answer 1

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 $n=1$.

2. 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.

Answer 2

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

Answer 3

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

Answer 4

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.

Answer 5

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 $\,n$.