<s>[INST] How prove this $\sum_{k=1}^{n}\frac{k}{3^k-2^k}<\frac{5}{3}$ prove that
$$\sum_{k=1}^{n}\dfrac{k}{3^k-2^k}<\dfrac{5}{3}$$
my idea: use 
$$3^k-2^k>2^k(k\ge 2)$$
then
$$\Longleftrightarrow 1+\sum_{k=2}^{n}\dfrac{k}{3^k-2^k}<1+\sum_{k=2}^{n}\dfrac{k}{2^k}<1+\sum_{k=2}^{\infty}\dfrac{k}{2^k}=1+\dfrac{3}{2}>\dfrac{5}{3}$$
use this methods,I can't prove it
other idea:
$$3^k-2^k>2\cdot 2^k(k\ge 3)$$
But This same can't prove it
$$\Longleftrightarrow \sum_{k=3}^{\infty}\dfrac{k}{3^k-2^k}<\sum_{k=3}^{\infty}\dfrac{k}{2^k}<\dfrac{2}{3}-\dfrac{2}{5}$$
so This problem have other nice methods? Thank you
 [/INST] By  arithmetic/geometric means, 
$$
3^k-2^k=3^{k-1}+3^{k-2}\cdot 2+\ldots +2^{k-1}< k\cdot 6^{(k-1)/2}.
$$
Hence
$$
\sum_{k\ge m}\frac{k}{3^k-2^k}<\sum_{k\ge m}\frac{1}{6^{(k-1)/2}}<\frac{1}{6^{(m-1)/2}}\frac{1}{1-\frac{1}{\sqrt{6}}}
$$
and it remains to choose $m$ as Raymond Manzoni advices.
</s>