Help Convergence test for series

[Puppy27]

Find the value of x so that the series below converge.[/b]

[tex]\sum [/tex] 1/ [(k^x) * (2^k)] (k=1 to [tex]\infty[/tex])
Using ratio test, I 've got
[(1/2) * 1^x] < 1 for all x in R
But when I use different value of x, series converge and diverge!

Really need your help! Thank you so much

 

[S_klogW]

I think we could say that for every x, the series can be convergent. But it's not uniformly convergent.
For all x ∈R，it's convergent. Just notice that if we assume S(x)=∑1/ [(k^x) * (2^k)] , we could find an integral number n with 1/ [(k^x) * (2^k)] ＜[(k^n) / (2^k)] .
let Sn=Σ[(k^n) / (2^k)], you could easily calculate Sn-0.5Sn, then for every n, the series can be convergent. But, unfortunately, the speed of the convergency depends on the integral n, which is based on x, so we could not say that the series uniformly converge.

 

[Puppy27]

That is why my lecturer asks to find values of x (or I think the range of value of x) to make this series converge

Help me please!

 

[Gib Z]

The ratio of consecutive terms is [tex] \frac{ k^x 2^k}{(k+1)^x 2^{k+1}} = \frac{1}{2} \cdot \left( \frac{k}{k+1} \right)^x [/tex].

You were incorrect to simply replace the k/(k+1) term with 1 straight away. You have to consider how different x values affect the term by breaking it up into cases. What if x is negative?

 

[Puppy27]

sorry, but the ratio test is using limit of U_k+1/U_k when k goes to infinity, right?
So that is why I get (1/2) * 1^x which is (1/2) for all x

 

[Gib Z]

So what is the problem? You said that when you test different values for x, some diverge and some converge. Which ones do you think make it diverge? You know that it must converge, so try to see where your reasoning is faulty.

 

[Puppy27]

sorry, I mistake the definition of convergence! I first thought a series converge when k increases, the value of each component in the series will increase.

so sorry!