<s>[INST] Limit of the function series I have to find the value
$$\lim_{k \to \infty}\sum_{n=1}^{\infty}\frac{k^3+n^3}{k^32^n+1}$$
Hint I've got:
Compare $$\frac{k^3+n^3}{k^32^n+1}$$ with $$\frac{1+n^3}{2^n}$$ and use Weierstrass theorem.
I do not know how to apply Weierstrass theorem with the limit of k.
Help guys :)
 [/INST] You have $\frac{k^3 + n^3}{k^3 2^n +1} -\frac{1}{2^n} = \frac{n^3 - 2^{-n}}{k^3 2^n +1}$ (and $\sum \limits_{n\geqslant 1} \frac{n^3 - 2^{-n}}{k^3 2^n +1}$ converges because the general term is equivalent to $\frac{n^3}{k^3 2^n}$).
So $\sum \limits_{n=1}^{\infty} \frac{k^3 + n^3}{k^3 2^n +1} = \sum \limits_{n=1}^{\infty} \frac{1} {2^n} + \frac{1}{k^3} \cdot \sum \limits_{n=1}^{\infty} \frac{n^3 - 2^{-n}}{2^n + \frac{1}{k^3}}$, and all the sums converge.
So finally the limit seeked is $\sum \limits_{n=1}^{\infty} \frac{1}{2^n} = 1$
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