# Show that $a_n$ is monotonic and bounded

I have the following sequence $$a_n=\frac{1}{2+3}+\frac{1}{2^2+3^2}+\frac{1}{2^3+3^3}+\ldots+\frac{1}{2^n+3^n}$$ and I need to show that it is monotonic and bounded.

It is easy to show that the sequence is increasing since $$a_{n+1}-a_{n}=\frac{1}{2^{n+1}+3^{n+1}}>0$$ for all $$n \in \mathbb{N}$$.

The sequence is also bounded from below because $$a_n \geq a_1=\frac{1}{5}$$.

I do not know how to find an upper bound. I would be grateful for any help.

• Compare with a geometric sum? – Mark Bennet Jan 11 at 18:02
• I wanted to bound this by $a_n<\frac{1}{2}+\frac{1}{2^2}+\ldots+\frac{1}{2^n}<\sum_{k=1}^{\infty}\frac{1}{2^k}$. Is it also correct? – Uhans Jan 11 at 18:09
• @Uhans Yes it is correct, my bound was wrong actually 😅 But the idea is the same – Maximilian Janisch Jan 11 at 18:10
• Ok! Thank you! :) – Uhans Jan 11 at 18:13

The monotony is as you said. The sequence is also bounded from above because, for each $$n$$ you have $$\frac{1} {{2^n + 3^n }} \leqslant \frac{1} {{2^n }}$$ and therefore $$a_n \leqslant \frac{1} {2} + \frac{1} {{2^2 }} + \cdots \frac{1} {{2^n }} \leqslant \sum\limits_{n = 0}^\infty {\frac{1} {{2^n }}} = 2$$