Question

Evaluate \[\sum_{k=0}^{\infty} \frac{2k + 1}{(4k + 1)(4k+3)(4k+5)}\]

Answer 1

How are you gonna do it without, Mathematica or Wolfram Alpha?

Answer 2

Use partial fractions to break the fraction up.

Answer 3

I am trying partial fractions :-/

Answer 4

Is that the only way?

Answer 5

most likely yes. It might turn into a telescoping sum.

Answer 6

i would like to try this problem, but my nails aren't dry yet. so its hard to type

Answer 7

umm I am getting

\[\sum_{k=0}^{\infty}\frac{2k+1}{16}\times\left(\frac{1}{4k+1} + \frac{2}{4k+3} - \frac{3}{4k+5} \right)\]

Answer 8

im getting this:\[\frac{2k+1}{(4k+1)(4k+3)(4k+5)}=\frac{1}{16}\left(\frac{1}{4k+1}\right)+\frac{1}{8}\left(\frac{1}{4k+3}\right)-\frac{3}{16}\left(\frac{1}{4k+5}\right)\]which isnt helping too much... i cant really afford to spend too much time on this as i have my own finals to study for, but here are some small things that may or may not lead to a solution:

Note that:\[\sum_{k=0}^{\infty}\frac{1}{4k+1}=1+\sum_{k=0}^{\infty}\frac{1}{4k+5}\]

Also note that:\[\frac{(4k+1)+(4k+5)}{2}=4k+3\]

Answer 9

Thanks Joe! and good luck for your exams!

Answer 10

i wish it was a telescoping series
:(
What ways have you learned ishaan?

Answer 11

i don't know, a friend of mine gave me this problem; i only know the basic stuff arithematic, geometric and harmonic, series of square and cubes.