Calculus- P-series test: Inf Sum n=1 [n/(8+n^2)^p

[Abraham]

Find the positive values of p for which the series converges. (Enter your answer using interval notation.)

Inf Sum n=1 [n/(8+n²)^p

I would really appreciate if someone can help me with this problem
I keep getting (1/2, inf) as the final answer but its wrong.

 

[Deleted member 4993]

Find the positive values of p for which the series converges. (Enter your answer using interval notation.)

Inf Sum n=1 [n/(8+n²)^p

I would really appreciate if someone can help me with this problem
I keep getting (1/2, inf) as the final answer but its wrong.

n/(8+n²)^P

= n^(1-2P)/[(1 + 8/n²)^P ]

Continue....

 

[stapel]

Find the positive values of p for which the series converges. (Enter your answer using interval notation.)

Inf Sum n=1 [n/(8+n²)^p

In mathematics, "Inf" typically stands for "infimum". However, based on what you say later, I think you mean "infty" or "infinity", and perhaps you're using an analog of the "m-choose-n" notation to indicate the upper and lower limits of a summation. However, I can't guess where the unbalanced square bracket belongs...? Do you mean either of the following?

. . . . .\(\displaystyle \displaystyle \mbox{a. }\, \sum_{n\, =\, 1}^{\infty}\, \)\(\displaystyle \left[\, \dfrac{n}{8\, +\, n^2}\, \right]^p\)

. . . . .\(\displaystyle \displaystyle \mbox{b. }\, \sum_{n\, =\, 1}^{\infty}\, \)\(\displaystyle \dfrac{n}{(8\, +\, n^2)^p}\)

When you reply, please include the expected answer, along with a clear listing of your steps so far, showing how you arrived at your own result. Thank you!

 

[Ishuda]

n/(8+n²)^P

= n^(1-2P)/[(1 + 8/n²)^P ]

Continue....

I would have written it differently [because I tend to think of n in the denominator for the p test] but would have had basically the same thing.

Anyway, the real reason for commenting is a fact I found out just recently: If you take the harmonic series [\(\displaystyle \sum_1^{\infty}\, \frac{1}{n}\)] but remove all n's with a 9 in it, the series converges
https://en.wikipedia.org/wiki/Kempner_series
In fact, if you remove any digit string, i.e. any n which has the digit string 31415 in it, the series converges. Now ain't that a hoot.