# Can someone explain to me why this series diverges?

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).

• 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 4 '14 at 6:33
• 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 4 '14 at 6:38
• Now, you don't have the variable $a$ anywhere in the summation. – JimmyK4542 Sep 4 '14 at 6:39
• @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 4 '14 at 6:40
• Is $n$ a constant? Or did you mean $$\sum_{k=1}^\infty \dfrac{1}{k^{3+\cos(k)}}$$ – Robert Israel Sep 4 '14 at 6:41

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.
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 $\frac{1}{k^{2+\cos k}}$.
• 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 4 '14 at 7:08