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February 14, 2020

Sequence and Series: IIT JAM 2016 Problem 24

Sequence and series of real numbers

A sequence of real numbers is an one to one mapping from $\mathbb{N}$ (the set of natural numbers) to a subset of $\mathbb{R}$(the set of all real numbers).

Sum of the terms of a real sequence sequence is called a series.

Try the problem

Find the sum of the series $\displaystyle\sum_{n=2}^{\infty} \frac{(-1)^n}{n^2+n-2}$.

$\textbf{(A)}\quad \frac13\ln 2-\frac{5}{18}\quad \textbf{(B)}\quad \frac13\ln 2-\frac{5}{6}\quad \textbf{(C)}\quad \frac23\ln 2-\frac{5}{18}\quad \textbf{(D)}\quad \frac23\ln 2-\frac{5}{6}\quad$


Sequence and Series of real numbers.

6 out of 10

Introduction to Real Analysis : Robert G. Bartle & Donald R. Sherbert.

Knowledge Graph

Sequence of real numbers-knowledge graph

Use some hints

This problem is a very basic one, this problem can easily be solved by step by step solution. The steps are:

Step 1 : First we will ignore the summation part. We will factorize the denominator, because we are going step by step so our aim is to simplify the given problem first.

Step 2: After factorizing the the denominator we will reach to a position where we have to use partial fraction to go forward.

Step 3: In this step we will take care of the $(-1)^n$ part, like how it will affect the series.

Step 4: After taking care of the $(-1)^n$ we will now expand the summation (breaking it into infinite sum).

Step 5 : So after $4$ steps we are halfway done now just the last simplification is left we will use the value

$\ln 2= 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots$

to simplify it further.

Now try to solve the entire problem by following these steps !!!

Let us see how to execute STEP 1 .

$\displaystyle\sum_{n=2}^{\infty} \frac{ (-1)^n }{ n^2+n-2 }$, here we are only concerned with the denominator part so,


So we have successfully factorized the denominator. Now you can reach to the part where we have to use the concept of partial fraction!!!

Let us execute STEP 2 now.

After factorizing the denominator in HINT 1 we get

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

Now our aim here is to seperate $(n+2)$ and $(n-1)$ so we can do this by using partial fraction.


$\Rightarrow A(n-1)+B(n+2)=1$

Now taking $\underline{n=1}$ we get



Again taking $\underline{n=-2}$, we get:


therefore, $A=-\frac{1}{3}$

So we get, $\frac{1}{(n+2)(n-1)}=\frac{1}{3}[\frac{1}{n-1}-\frac{1}{n+2}]$

So our aim here is successful we have separated $(n+2)$ and $(n-1)$ . Now can you proceed further ???

Now after using partial fraction we get.

$\displaystyle\sum_{n=2}^{\infty} (-1)^n\frac{1}{3}[\frac{1}{n-1}+\frac{1}{n+2}]$

Now we will execute STEP 4 i.e., we will take care of the $(-1)^n$ and at the same time we will execute STEP 4 i.e, we will split it into the infinite sum.

$\frac{1}{3}\displaystyle\sum_{n=2}^{\infty} [\frac{1}{n-1}+\frac{1}{n+2}](-1)^n$

$\Rightarrow \frac{1}{3}[(1-\frac{1}{4})-(\frac{1}{2}-\frac{1}{5})+(\frac13-\frac16)\ldots]$

$\Rightarrow \frac13[(1-\frac12+\frac13-\frac14\ldots)+(-\frac14+\frac15-\frac16\ldots)]$

Now after this can you applying the formula of '$\ln 2$' to finish the problem!!

Now only STEP 5 is left to execute. So we will use the infinite series of $\ln 2$.


$=\frac13[\ln 2+\ln 2-(1-\frac12+\frac13)]$

$=\frac13(2\ln 2-\frac56)$

$=\frac23 \ln 2 - \frac{5}{18}$

Hence the anser is C.

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