Verify if this $\sum_{n=1}\frac{n^3+a}{2^n+a}$ will be convergent or divergent?

Question

I already calculated the first fraction which is : $$\frac{n^3}{2^n+a}$$ as convergent for all values of a by using $\frac{n^3}{2^n}$ but I can't find a method to calculate the other one .

Is there any quick solution for the whole fraction instead of solving each part at a time?

Answer 1

If you proved that the first sum is convergent than also the second one is convergent, because:

$$\exists N:\frac{a}{2^n+a}<\frac{n^3}{2^n+a} \ \ \ \forall n\geq N$$

So for comparison test the sum converges. Clearly this is valid only if $a\neq -2^k \ \ \ k\in \Bbb{N}-\{0\}$

:)

Answer 2

We have $n^3+a\le2n^3$ for $n$ large enough (i.e $n\ge a^{1/3}$).

We have also $2^n+a\ge 2^{n/2}$ for $n$ large enough (i.e $n\ge 2\log_2|a|$ and $n\ge 2$)

So $0\le a_n=\dfrac{n^3+a}{2^n+a}\le b_n=\dfrac{2n^3}{2^{n/2}}$

We can apply the ratio test and $\left|\dfrac{b_{n+1}}{b_n}\right|=\dfrac{2(n+1)^3\times 2^{n/2}}{2n^3\times 2^{(n+1)/2}}=\dfrac{n+1}{n\sqrt{2}}\to \dfrac 1{\sqrt{2}}<1$

Thus $\sum b_n$ is absolutely convergent and so does $\sum a_n$ by the comparison test.

Note: for $a$ not a negative power of $2$ obviously

If you know of equivalents, you can argue directly that $a_n\sim\dfrac{n^3}{2^n}$ and conclude like previously rather than bounding numerator and denominator.