# Series Convergence

#### joshuaa

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

#### topsquark

Forum Staff
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

1 person

#### joshuaa

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.

1 person

#### joshuaa

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?

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#### Idea

$$\displaystyle 0<\frac{e^x}{\left(2+e^{2x}\right)^k}<\frac{e^x}{e^{2k x}}=\left(e^{1-2k}\right)^x$$

2 people

#### joshuaa

thanks Idea and topsquark.