 ×     #### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
```
1/1+1 2 +1 4 +1/1+2 2 +2 4 +1/1+3 2 +3 4 +......... infinite find 14s

```
3 years ago

```							The n th term in the series can be written as:S(n) = n/(1 + n² + )Our aim is to convert this into a form where we can cancel adjacent terms and finally reach at a numerical value.Therefore we should convert this into term1  -  term2 form , by using partial fractions.S(n) = n/(1 + n² + ) = n/(S(n) = n/ (n² + 1)² - n² =By applying a² - b² = (a + b)(a - b)s(n) = n/ (n² + n + 1)(n² - n + 1)Now we can writen = 1/2 (n² + n + 1) - (n² - n + 1)Substituting this in S(n),S(n) = 0.5 x  = 0.5 x  - Therefore,Now adding each terms for n = 1 , 2 , 3 , ....Sum = 0.5 ( 1 - 1/3 + 1/3 - 1/7 + 1/7 - 1/13 ......)We can observe that the last partial fraction of nth term and first partial fraction of n+1 th term gets cancelled.This can be extended to all n as it goes till infinity.Therefore ,Sum = 0.5Therefore the sum of the series till infinity is 0.5
```
7 months ago
Think You Can Provide A Better Answer ?

## Other Related Questions on IIT JEE Entrance Exam

View all Questions »  ### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution  ### Course Features

• 731 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions