I am trying to prove convergence of the series
$(1) \sum_{n=1}^\infty \frac{\sin(n)}{\sqrt{n}+\sin(n)}$.
If this series were $(2) \sum_{n=1}^\infty \frac{\sin(n)}{\sqrt{n}}$, then it converges by the Dirichlet test because partial sums of $\sin(n)$ are bounded and $1/\sqrt{n}$ decreases monotonically to $0$.
The difficulty I have is that $\sqrt{n}+ \sin(n)$ in the denominator is not monotonic and $\sin(n)$ in the numerator changes sign. But the denominator is bracketed by monotonic sequences $\sqrt{n} – 1 \leq \sqrt{n} + \sin(n) \leq \sqrt{n} + 1$ and it seems possible to adapt the Dirichlet test.
Can this be done to prove convergence of (1) or is there a more general approach that I am not aware of?
Best Answer
Let me help you with this one.
We can rewrite the following expression as follows: $$ \sum_{n=1}^\infty {\sin(n)}\cdot(\sin(n)+n^{1/2})^{-1} \tag{1} $$ then we can take $n^{1/2}$ out and write it like $$ \sum_{n=1}^\infty {\frac{\sin(n)}{n^{1/2}}}\cdot\left (1+\frac{\sin(n)}{n^{1/2}}\right)^{-1}\tag{2} $$ now we can apply the taylor series like $$ \sum_{n=1}^\infty {\frac{\sin(n)}{n^{1/2}}}\cdot\left(1-\frac{\sin(n)}{n^{1/2}} + O\left(\frac{1}{n}\right)\right)\tag{3}$$
from here you can solve it, first one converges by Dirichlet, second one diverges, third one converges.