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Convergence:

$$\sum^\infty_{n=0}\frac{\cos \left(n + \frac{1}{n^2} \right)}{n \cdot \ln \left( n^2 + 1 \right)}$$

Edit (I tried to follow the idea written by @Daniel Fischer):

$$\sum_{n=0}^{\infty} \frac{cos(n + \frac{1}{n^2})}{n\cdot ln(n^2 + 1)} = \sum_{n=0}^{\infty} \frac{cos(n)}{n \cdot ln(n^2 + 1)} + \sum_{n=0}^{\infty} \frac{cos(n + \frac{1}{n^2}) - cos(n)}{n \cdot ln(n^2 + 1)}$$

$\sum_{n=0}^{\infty} \frac{cos(n)}{n \cdot ln(n^2 + 1)}$ is convergent because of Dirichlet test and How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?

$\sum_{n=0}^{\infty} \frac{cos(n + \frac{1}{n^2}) - cos(n)}{n \cdot ln(n^2 + 1)} = \sum_{n=0}^{\infty} \frac{-2 \left(sin \left( \frac{2n+\frac{1}{n^2}}{2} \right) \cdot sin \left( \frac{\frac{1}{n^2}}{2} \right) \right)}{n \cdot ln \left(n^2 + 1 \right)} = \sum_{n=0}^{\infty} \frac{-2 \left(sin \left( n+\frac{1}{2n^2} \right) \cdot sin \left( \frac{1}{2n^2} \right) \right)}{n \cdot ln \left(n^2 + 1 \right)}$

Then: $$\sum_{n=0}^{\infty} \left| \frac{-2 \left(sin \left( n+\frac{1}{2n^2} \right) \cdot sin \left( \frac{1}{2n^2} \right) \right)}{n \cdot ln \left(n^2 + 1 \right)} \right| \leq \sum_{n=0}^{\infty} \left| 2 \cdot \frac{ \frac{1}{2n^2} }{n \cdot ln \left(n^2 + 1 \right)} \right| \leq \sum_{n=0}^{\infty} \left| \frac{ 1 }{n^3 \cdot ln \left(n^2 + 1 \right)} \right| $$

For $n \geq 3$: $$\sum_{n=0}^{\infty} \left| \frac{ 1 }{n^3 \cdot ln \left(n^2 + 1 \right)} \right| \leq \sum_{n=0}^{\infty} \left| \frac{1}{n^2} \right|$$

Therefore we know that $\sum_{n=0}^{\infty} \frac{cos(n + \frac{1}{n^2}) - cos(n)}{n \cdot ln(n^2 + 1)}$ it is also convergent. Is that correct?

mymathc
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    Write $$\cos\biggl(n + \frac{1}{n^2}\biggr) = \cos n + \biggl(\cos \biggl(n + \frac{1}{n^2}\biggr) - \cos n\biggr),.$$ – Daniel Fischer Jan 06 '21 at 21:36
  • @Daniel Fischel could you tell me if I did it correct? I tried to use your hint. – mymathc Jan 07 '21 at 01:57
  • I see now (overlooked it yesterday) that you start your summation at $n = 0$, but the term for $n = 0$ is undefined. Start the summation at $n = 1$ (or later). Apart from that, everything is correct. At the end, the $n \geq 3$ is a little odd (I guess it comes from $\ln n \geqslant 1$ for $n \geqslant 3$, but here we could use $\ln (n^2+1) > 1$ for $n \geqslant 2$). Also that you use $\frac{1}{n^2}$ for the comparison rather than $\frac{1}{n^3}$ is a little odd. I have two ideas why you did that, a) you prefer to use a more "famous" series for comparison, – Daniel Fischer Jan 07 '21 at 11:43
  • or b) you prefer to use a weaker result when that suffices. Both are valid and often useful strategies, but if not made explicit can lead to the question "Why did the author not use [whatever] here?". Depending on the background, it can be simpler/preferable to use the mean value theorem to estimate the difference of the cosines rather than the addition theorem plus $\lvert \sin x\rvert \leqslant \lvert x\rvert$ as you did. It's less to type at least. But the addition theorem may be available before differentiation, so that's (possibly) a more elementary way. – Daniel Fischer Jan 07 '21 at 11:43

1 Answers1

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Continuing from the remark made in the comments, $$\cos\biggl(n + \frac{1}{n^2}\biggr) = \cos n + \biggl(\cos \biggl(n + \frac{1}{n^2}\biggr) - \cos n\biggr).$$ The second term on the right decays like $1/n^2$, since cosine is Lipschitz continuous (prove this using the cosine sum identity). The partial sums of the first term on the right are uniformly bounded; to see this, it is convenient to remember cosine is the real part of the complex exponential, $$ \cos n = \operatorname{Re} e^{in}. $$ Thus the partial sum of cosine is just the real part of a nice geometric series, $$ \sum_{n = 0}^m \cos n = \operatorname{Re} \sum_{n = 0}^m e^{in} = \operatorname{Re} \frac{1 - e^{i(m + 1)}}{1 - e^i}. $$ The right hand side has uniformly bounded modulus in $m$, so we are done.

  • I like that solution. Could you check mine? I will mark your answer as the correct one if everyting is ok. – mymathc Jan 07 '21 at 01:35
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    @mymathc yes your solution is essentially equivalent to mine, you implicitly proved Lipschitz continuity of cosine in your solution. –  Jan 07 '21 at 04:23
  • Thank you, finally, that makes me happy :D – mymathc Jan 07 '21 at 04:24