It is easy to see that $$\sum_{n = 10}^\infty \frac{\sin n}{n + 10 \sin n} = K + \sum_{n = 11}^\infty \frac{\sin n}{n + 10 \sin n} \le K + \sum_{n=11}^\infty \frac{\sin n}{n - 10},$$

where $$K = \sum_{n = 10}^{10} \frac{\sin n}{n + 10\sin n} = \frac{\sin 10}{10 + 10\sin10} \in \mathbb{R}.$$

(This is to avoid the undefined term $\frac{\sin 10}{10 - 10}$ in the upper bound.)

Let's write $a_n := \sin n$ and $b_n := \frac1{n - 10}$. The sequence of partial sums of $a_n$ is bounded and $b_n$ is a monotonic sequence that tends towards $0$ as $n$ tends to infinity.

According to Dirichlet test, the series $\sum_{n=11}^\infty \frac{\sin n}{n - 10}$ converges. Accoding to the comparison test, the series $\sum_{n = 11}^\infty \frac{\sin n}{n + 10 \sin n}$ converges as well. Then the given series $\sum_{n = 10}^\infty \frac{\sin n}{n + 10 \sin n}$ is convergent, too.

Is my argumentation correct?

Thank you!

  • $\begingroup$ I don't understand: what undefined term in what upper bound? Certainly $\,\sin 10\neq -1\;$ ... $\endgroup$ – DonAntonio Aug 20 '13 at 13:21
  • $\begingroup$ Comparison test...between what and what? Remember this test applies for positive series... $\endgroup$ – DonAntonio Aug 20 '13 at 13:22
  • $\begingroup$ I will try to explain myself better: $K + \sum_{n = 11}^\infty \frac{\sin n}{n - 10}$ is the upper bound; now for $n = 10$, you get $10 - 10$ in the denominator. $\endgroup$ – David Aug 20 '13 at 13:24
  • $\begingroup$ @DonAntonio That is a bug. Thanks. $\endgroup$ – David Aug 20 '13 at 13:26
  • $\begingroup$ Ok, I'm confused: you upper-bound your series by $\,K+\sum_{n=11}^\infty\frac{\sin n}{n-10}\;$...so? If your series is not positive you can't use the comparison test. If you want to can try with absolute values and absolute convergence... $\endgroup$ – DonAntonio Aug 20 '13 at 13:27

Your approach is a bit awkward. The simplest approach would be to compare your $n$th term to $\frac{\sin n}{n}$ by showing that the difference $\frac{\sin n}{n + 10 \sin n}-\frac{\sin n}{n}$ is dominated by $\frac{1}{n^2}$ which is absolutely convergent. The convergence of your series then follows from the convergence of the classical series $\sum\frac{\sin n}{n}$.

  • $\begingroup$ Very nice. Thank you! $\endgroup$ – David Aug 20 '13 at 13:40

That is an interesting proof. I am going to generalize it by showing that, if $f(n)$ is a bounded function such that $\sum_{n=1}^{\infty} \frac{f(n)}{n}$ converges and $a$ is a real number such that $n+a f(n) \ne 0$ for all $n$, then $\sum_{n=1}^{m} \frac{f(n)}{n+a f(n)}$ converges as $m \to \infty$.

$a=10$ and $f(n) = \sin(n)$ in the original problem.

Consider $\sum_{n=1}^{m} \frac{f(n)}{n+a f(n)}$.

$\begin{align} \frac{f(n)}{n+a f(n)} -\frac{f(n)}{n} &=f(n)(\frac1{n+a f(n)}-\frac1{n})\\ &=f(n)\frac{n-(n+af(n))}{n(n+af(n))}\\ &=f(n)\frac{-af(n))}{n(n+af(n))}\\ &=\frac{-af^2(n))}{n(n+af(n))}\\ \end{align} $

so that $\sum_{n=1}^m \frac{f(n)}{n+a f(n)} =\sum_{n=1}^m \frac{f(n)}{n} +\sum_{n=1}^m \frac{-af^2(n))}{n(n+af(n))} $.

Since, as $m \to \infty$, $\sum_{n=1}^m \frac{f(n)}{n}$ converges (by assumption) and $\sum_{n=1}^m \frac{-af^2(n))}{n(n+af(n))}$ converges (by comparison with $\sum_{n=1}^m \frac1{n^2}$), $\sum_{n=1}^m \frac{f(n)}{n+a f(n)}$ also converges.

If $n+a f(n) = 0$ for some $n$, the result still holds if we start the sums at an $k$ such that $k > |a| M$, where $M> |f(n)|$ for all $n$ ($M$ exists since $f$ is bounded).


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