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.
Find the sum of the series $\displaystyle\sum_{n=2}^{\infty} \frac{(-1)^n}{n^2+n-2}$.
$\textbf{(A)}\quad \frac13\ln 2-\frac{5}{18}\quad \textbf{(B)}\quad \frac13\ln 2-\frac{5}{6}\quad \textbf{(C)}\quad \frac23\ln 2-\frac{5}{18}\quad \textbf{(D)}\quad \frac23\ln 2-\frac{5}{6}\quad$
IIT JAM 2016, PROBLEM 24
Sequence and Series of real numbers.
6 out of 10
Introduction to Real Analysis : Robert G. Bartle & Donald R. Sherbert.
First hint
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 !!!
Second Hint
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,
$\frac{1}{n^2+n-2}=\frac{1}{(n+2)(n-1)}$
So we have successfully factorized the denominator. Now you can reach to the part where we have to use the concept of partial fraction!!!
Third Hint
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.
$\frac{1}{(n+2)(n-1)}=\frac{A}{n+2}+\frac{B}{(n-1)}$
$\Rightarrow A(n-1)+B(n+2)=1$
Now taking $\underline{n=1}$ we get
$A(1-1)+B(1+2)=1$
$B=\frac{1}{3}$
Again taking $\underline{n=-2}$, we get:
$A(-2-1)+B(-2+2)=1$
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 ???
Fourth Hint
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!!
Final Step
Now only STEP 5 is left to execute. So we will use the infinite series of $\ln 2$.
$\frac13[(1-\frac12+\frac13-\frac14\ldots)+(-\frac14+\frac15-\frac16\ldots)]$
$=\frac13[\ln 2+\ln 2-(1-\frac12+\frac13)]$
$=\frac13(2\ln 2-\frac56)$
$=\frac23 \ln 2 - \frac{5}{18}$
Hence the anser is C.
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.
Find the sum of the series $\displaystyle\sum_{n=2}^{\infty} \frac{(-1)^n}{n^2+n-2}$.
$\textbf{(A)}\quad \frac13\ln 2-\frac{5}{18}\quad \textbf{(B)}\quad \frac13\ln 2-\frac{5}{6}\quad \textbf{(C)}\quad \frac23\ln 2-\frac{5}{18}\quad \textbf{(D)}\quad \frac23\ln 2-\frac{5}{6}\quad$
IIT JAM 2016, PROBLEM 24
Sequence and Series of real numbers.
6 out of 10
Introduction to Real Analysis : Robert G. Bartle & Donald R. Sherbert.
First hint
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 !!!
Second Hint
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,
$\frac{1}{n^2+n-2}=\frac{1}{(n+2)(n-1)}$
So we have successfully factorized the denominator. Now you can reach to the part where we have to use the concept of partial fraction!!!
Third Hint
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.
$\frac{1}{(n+2)(n-1)}=\frac{A}{n+2}+\frac{B}{(n-1)}$
$\Rightarrow A(n-1)+B(n+2)=1$
Now taking $\underline{n=1}$ we get
$A(1-1)+B(1+2)=1$
$B=\frac{1}{3}$
Again taking $\underline{n=-2}$, we get:
$A(-2-1)+B(-2+2)=1$
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 ???
Fourth Hint
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!!
Final Step
Now only STEP 5 is left to execute. So we will use the infinite series of $\ln 2$.
$\frac13[(1-\frac12+\frac13-\frac14\ldots)+(-\frac14+\frac15-\frac16\ldots)]$
$=\frac13[\ln 2+\ln 2-(1-\frac12+\frac13)]$
$=\frac13(2\ln 2-\frac56)$
$=\frac23 \ln 2 - \frac{5}{18}$
Hence the anser is C.