Series Convergence

Mar 2012
588
32




What are the values of k that make the series convergent?
 

topsquark

Forum Staff
Jan 2006
11,602
3,458
Wellsville, NY
Just a thought, give this a try: Let \(\displaystyle u = 2 + e^{2x}\). Then we have that \(\displaystyle e^x = \sqrt{u - 2}\).

Then your sum becomes
\(\displaystyle \sum_{k = 1}^{\infty} \dfrac{\sqrt{u - 2}}{u^k}\)

which looks a lot simpler.

-Dan
 
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Mar 2012
588
32
Just a thought, give this a try: Let \(\displaystyle u = 2 + e^{2x}\). Then we have that \(\displaystyle e^x = \sqrt{u - 2}\).

Then your sum becomes
\(\displaystyle \sum_{k = 1}^{\infty} \dfrac{\sqrt{u - 2}}{u^k}\)

which looks a lot simpler.

-Dan

I want x to go from 1 to infinite, not k. I want to figure what are the values of k.
For example, when k = 0, the series diverges, but k = 1, it converges.
 
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Mar 2012
588
32
by plugging random values of k, I observed that the sum increases very very slightly like it will reach finite value for k > 0.5.

If k <= 0.5, the sum will increase gradually to infinite.

Does that mean the series is converges when k > 0.5?
How to prove that mathematically?
 
Last edited:
Jun 2013
1,151
614
Lebanon
\(\displaystyle 0<\frac{e^x}{\left(2+e^{2x}\right)^k}<\frac{e^x}{e^{2k x}}=\left(e^{1-2k}\right)^x\)
 
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