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I have this series

$$\sum_{k=1}^{\infty}\frac{1}{k^{3+\cos k}}$$

I understand that if the exponent is fixed (not a function) and greater than 1 the series converges (p-series) but I can see in wolfram that this series diverges clearly (wolfram says that the comparison test shows that the series diverges... but I dont know what series is using on the comparison).

Did
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Masacroso
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  • Just to be clear, did you mean for the summand to be $\dfrac{1}{a^{3+\cos k}}$ instead of $\dfrac{1}{a(3+\cos k)}$? – JimmyK4542 Sep 04 '14 at 06:33
  • The first. Sry, I did some mistakes. I put a but I was talking about harmonic-type series and not geonetric-type derivations. – Masacroso Sep 04 '14 at 06:38
  • Now, you don't have the variable $a$ anywhere in the summation. – JimmyK4542 Sep 04 '14 at 06:39
  • @JimmyK4542, yes, sorry for the first mistakes... No a, just a derivation of Riemann series with function on exponent. Sry first mistakes. If I put a it isnt what I was trying to compare. – Masacroso Sep 04 '14 at 06:40
  • Is $n$ a constant? Or did you mean $$\sum_{k=1}^\infty \dfrac{1}{k^{3+\cos(k)}}$$ – Robert Israel Sep 04 '14 at 06:41
  • $$\frac{1}{k^{3+\cos k}}\leq\frac{1}{k^2}$$ so convergent, no? – JP McCarthy Sep 04 '14 at 06:42
  • @RobertIsrael,@Jp McCarthy GOD, sry guys. Yes, is k, I fixed it. Sry. Now is the correct version what Im trying to know. Sry. – Masacroso Sep 04 '14 at 06:42
  • Then the comparison $$\frac{1}{k^{3+\cos k}}\leqslant\frac1{k^2}$$ shows that the series converges, not that it diverges. – Did Sep 04 '14 at 06:45
  • Well if we have finally got to $\sum\frac{1}{k^{3+\cos k}}$, that obviously converges. – almagest Sep 04 '14 at 06:45
  • Please do not delete the considerations related to the Wolfram program. – Did Sep 04 '14 at 06:48
  • Ok, thank you guys. I get a bit confused trying to understand this question. – Masacroso Sep 04 '14 at 06:51
  • @Did I deleted because I get confused with a different function. I was seeing various sum on wolfram. The consideration that I deleted was about a different sum not the sum on the question. Sry. – Masacroso Sep 04 '14 at 06:52

2 Answers2

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This series is convergent. Since $\cos k\ge -1$, $$\frac{1}{k^{3+\cos k}}\le \frac {1}{k^2}.$$ It would be more interesting if it were $\frac{1}{k^{2+\cos k}}$.

Quang Hoang
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  • Can you extend on the example with 2 instead of 3? Please? I will appreciate. – Masacroso Sep 04 '14 at 06:53
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    Masacroso: You should be aware now that the series of general term $1/k^{2+\cos k}$ is an entirely different story... hence asking about it on the present page is inappropriate. – Did Sep 04 '14 at 07:08
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Since $3 + \cos(k) \ge 2$ for all $k$, $\dfrac{1}{k^{3+\cos(k)}} \le \dfrac{1}{k^2}$, so this series converges by the comparison test. You might have confused Wolfram by using $n$ instead of $k$: if $n$ is a constant the series $\sum_k \dfrac{1}{n^{3+\cos(k)}}$ diverges because the terms are bounded below.

Robert Israel
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  • Yes, surely it was some confuse thing that I did. I was seeing various sums on wolfram and I surely "mixed" things of differents ones. Sry. – Masacroso Sep 04 '14 at 06:55