The series 1 n-1 - 1 2 n-1 2 - 1 3 n-1 3 - has the same sum as the series

The correct option is A 1 n 1 2 n ^ 2 + 1 3 n ^ 3 1 4 n ^ 4 +⋯ Given, 1 ( n+1 ) nbsp; + 1 2( n+1 ) nbsp;^ 2 + 1 3( n+1 ) nbsp ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Sum of Product of Binomial Coefficients

The series ...

Question

The series $\frac{1}{(n + 1)} + \frac{1}{2 {(n + 1)}^{2}} + \frac{1}{3 {(n + 1)}^{3}} + \dots$ has the same sum as the series

\frac{1}{n} - \frac{1}{2 n^{2}} + \frac{1}{3 n^{3}} - \frac{1}{4 n^{4}} + \dots

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{1}{n} + \frac{1}{2 n^{2}} + \frac{1}{3 n^{3}} + \frac{1}{4 n^{4}} + \dots

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{n} + \frac{1}{2^{2}} \cdot \frac{1}{n^{2}} + \frac{1}{2^{3}} \cdot \frac{1}{n^{3}} + \dots

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of the above

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{1}{n} - \frac{1}{2 n^{2}} + \frac{1}{3 n^{3}} - \frac{1}{4 n^{4}} + \dots$
Given, $\frac{1}{(n + 1)} + \frac{1}{2 {(n + 1)}^{2}} + \frac{1}{3 {(n + 1)}^{3}} + \dots$
$= - {- \frac{1}{(n + 1)} - \frac{1}{2 {(n + 1)}^{2}} - \frac{1}{3 {(n + 1)}^{3}} - \dots \infty}$
$= - log {1 - \frac{1}{(n + 1)}}$
$= - log {\frac{n}{n + 1}}$
$= log (\frac{n + 1}{n}) = log (1 + \frac{1}{n})$
$= \frac{1}{n} - \frac{1}{2 n^{2}} + \frac{1}{3 n^{3}} - \frac{1}{4 n^{4}} + \dots$

Suggest Corrections

Similar questions

$\frac{1}{n + 1} + \frac{1}{2 (n + 1)^{2}} + \frac{1}{3 (n + 1)^{3}} + . . .$ = a $\times$ $\frac{1}{n} - \frac{1}{2 n^{2}} + \frac{1}{3 n^{3}} - . . .$ Find a

Q. Show that the series whose

n

th term is

\frac{\sqrt[3]{2 n^{2} - 1}}{\sqrt[4]{3 n^{3} + 2 n + 5}}

is divergent.

Q. The sum of the series

2 C_{0} + \frac{C_{1}}{2} \cdot 2^{2} + \frac{C_{2}}{3} \cdot 2^{3} + \frac{C_{3}}{4} \cdot 2^{4} + \dots + \frac{C_{n}}{n + 1} \cdot 2^{n + 1}

is equal to , where

(C_{r} =^{n} C_{r})

{lim}_{n \to \infty} (\frac{n}{n^{2}} + \frac{n}{n^{2} + 1} + \frac{n}{n^{2} + 2^{2}} + . . . . . . + \frac{n}{2 n^{2} - 2 n + 1})

is equal to

{lim}_{n \to \infty} \frac{1. n^{2} + 2 (n - 1)^{2} + 3 (n - 2)^{2} + . . . + n {.1}^{2}}{1^{3} + 2^{3} + 3^{3} + . . . + n^{3}}

Join BYJU'S Learning Program

The series 1(n+1)+12(n+1)2+13(n+1)3+⋯ has the same sum as the series

The correct option is A 1n−12n2+13n3−14n4+⋯Given, 1(n+1)+12(n+1)2+13(n+1)3+⋯=−{−1(n+1)−12(n+1)2−13(n+1)3−⋯∞}=−log{1−1(n+1)}=−log{nn+1}=log(n+1n)=log(1+1n)=1n−12n2+13n3−14n4+⋯

The series $\frac{1}{(n + 1)} + \frac{1}{2 {(n + 1)}^{2}} + \frac{1}{3 {(n + 1)}^{3}} + \dots$ has the same sum as the series