- #1

- 4

- 0

[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

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- Thread starter Puppy27
- Start date

- #1

- 4

- 0

[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

- #2

- 19

- 0

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.

- #3

- 4

- 0

Help me please!!!

- #4

Gib Z

Homework Helper

- 3,346

- 5

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?

- #5

- 4

- 0

So that is why I get (1/2) * 1^x which is (1/2) for all x

- #6

Gib Z

Homework Helper

- 3,346

- 5

- #7

- 4

- 0

so sorry!

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