IIT JEE , AIEEE Forum - Mathematics , Algebra

coolvignesh

hey guys.....

can u plzz telll me hw to find sum to n terms of series

1/2 + 2/3 + 3/4 ..... n/n+1

computer001

http://www.goiit.com/posts/list/algebra-summation-of-series-48854.htm#245473

coolvignesh

so this series cant be done????????

computer001

thts how it seems...but more ppl will try..then v'll no..

deedee

S = 1/2 + 2/3 + 3/4 + .........n/(n+1)

Write S as (1-1/2) + (1-1/3) + (1-1/4)...... 1-1/(n+1)
S = n -(1/2 + 1/3 + 1/4 + ....1/n+1)

after this step

can we convert those termz in hp into ap n then finda summation

itz d summation of 2+3+4+5.....(n+1) termzzz

so itzn/2(n+1)+n

so d whole xpression bcumx

n/2(n+1)

cant we do it like that ??/

computer001

deedee...
it is humanly impossible to find the summation of an H.P

computer001

plz tell how from 1/2 + 1/3...
u got it to summation of 2+3+4...

pinnacle

1/2+ 1/3 +1/4....
is logarithm series i suppose :-s
m i doin sumthin wrong ?

Sushmi

Yes deedee u got it wrong

it will be 3.4.5.....+ 2.4.5......+2.3.5......+......like this it continues and

pinnacle exponential series is an infinite series.here series is not infinite.

It would be bettr to leave it.

pinnacle

yup
dats d doubt i had
it must be havin sum solution though

Also
sum of a hp is not reciprocal of the sum of ap
so whosoeva did dat
dats wrong buddy

deedee

yeah ................thanx 4 correctin

stupid em !!

djdylan2000

Ask the experts if they can do it.

rautela

s = 1/2 + 2/3 + 3/4........................... +n

s = 1/2 + 2/3 + 3/4 ...................+n-1+ n

s-s= 1/2+1/2+1/2.............................n -1 -n

n = 1/2+1/2+1/2.............................n -1

t(n)= 1/2((r^n)-1/2)/1/2

now this will be the nth term put summation frm n=1 to n

for term at n

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