Hi can someone help me find a formula for the nth term of telescoping series?? Then need to find limit. Equation enclosed.

Question

anonymous · Answer

$$\sum_{1}^{\infty} (1/(k+1)) - (1/(k+2))$$

amistre64 · Answer

if its telesoping, then certain pairs cancel out

amistre64 · Answer

write out a few terms if need be

anonymous · Answer

Yes so I will be left with first and last terms I believe. should be 1/2 - 1/k+2

amistre64 · Answer

seems about right

amistre64 · Answer

1/infinity = 0 doesnt it?

amistre64 · Answer

or rather it limits itslef to zero

anonymous · Answer

my book shows this and then says it is equal to k/2k+4

anonymous · Answer

which is th epart I am having trouble understanding. I took my limit with 1/2 - 1/k+2 which equals 1/2

anonymous · Answer

Bu tI dont undertstand how they come up with the generic formula which was supposed to be in terms of n ......n/2n+4

amistre64 · Answer

1        1
-  -  -----
2       k+2


(k+2)          2
-----  -   -----
2(k+2)    2(k+2)



   k
-----
2k+4

anonymous · Answer

ahhhh thanks (;

amistre64 · Answer

just add the fractions

amistre64 · Answer

since the top and bottom are of the same degree ... it limits to 1/2 as well

amistre64 · Answer

but that could be seen simply from the first setup.

1        1
-  -  -----
2       k+2 
       ^^^^^^
         this term limits to zero leaving us with 1/2

anonymous · Answer

right.....so this is a good formula for the nth term but I like the other version better with first and last term to find the limit ........

anonymous · Answer

exactly...

anonymous · Answer

thank you

amistre64 · Answer

youre welcome :)