 ×     #### 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
```
The sum to infinity of the series 1/1 + (1/1+2) + (1/1+2+3)+...... is equal to...

```
3 years ago

```							The general term of the series is 1/[r(r+1)/2]= 2/[r(r+1)]. It can be written as 2[(1/r)-(1/r+1)], so for calculating sum put n in place of r therefore the answer is [2n/(n+1)]
```
3 years ago
```							Ww can write ggeneral term Tn Tn = 1/[n(n+1)/2]Tn = 2/n(n+1)We can simplify it by partial fractionTn = 2[1/n - 1/(n+1)]Now we have to calculate the sum to infinite termsT1 +T2 +T3+ T4+.......= 2[1/1 - 1/2] + 2[1/2 - 1/3] +.....so onHence every terms cancel except the first termsHence sum  = 2 *1/1 = 2Hope it helps
```
3 years ago
```							t’s actually simple….Let series S= 1/1+1/3+1/6+1/10+1/15+…….S can be rewritten asS = 2 * (1/2+1/6+1/12+1/20+1/30+……..)Now, let’s consider each term1/2 = 1 - 1/21/6 = 1/2 - 1/31/12 = 1/3 - 1/41/20 = 1/4 - 1/51/30 = 1/5 - 1/6Thus,
```
2 months ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Algebra

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

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