Question

the sum upto infinity of 1/1+1/1+2+1/1+2+3......

Answer 1

\[\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}......\]

Answer 2

@UnkleRhaukus help

Answer 3

@ganeshie8 hlp

Answer 4

ok so next term is 1/1+2+3+4 right ?

Answer 5

can we write last terms as zero

Answer 6

Hint :-
\(1+2+3+... + n = \dfrac{n(n+1)}{2}\)

Answer 7

so the series we have would be like
\(\sum_{i=1}^{n} \dfrac {1}{\dfrac{i(i+1)}{2}}\)

Answer 8

so1/ n(n+1) will be zero

Answer 9

mm limit when n goes to infinity =0

Answer 10

answer is 2

Answer 11

ok i got it thanks @ikram002p

Answer 12

triangle numbers

Answer 13

it will be inverse of triangle number like this :-

\(2(\dfrac{1}{2} +\dfrac{1}{6} +\dfrac{1}{12} +\dfrac{1}{20} +...)\)

Answer 14

@kanwal32 np :)