The behaviour of $\sum k^a 2^k$.

Question

Let $a$ be a real number, I want to find a simple equivalent (or if it is possible, an asymptotic expansion) of $$\sum_{k=1}^n k^a 2^k.$$ I believe that the sum is $\sim n^a 2^{n+1}$ (tried many values of $a$ for large $n$). It is easy to prove it if $a\in \mathbb{N}$ by recurrence, but I don't know what to do if $n\notin \mathbb{N}$.

Answer 1

Let $S_{n,a}=\sum_{k=1}^n k^a 2^k$

• Prove first that $S_{n,a}=O(n^a 2^{n+1})$

If $a\geq 0$: $S_{n,a}\leq n^a\sum_{k=1}^n 2^k=n^a(2^{n+1}-2)\leq n^a 2^{n+1}$

If $a<0$, $S_{n,a}\leq \sum_{k=1}^{\lfloor n/2 \rfloor} k^a 2^k + \sum_{k=\lfloor n/2 \rfloor+1}^{n} k^a 2^k\leq 2^{\lfloor n/2 \rfloor+1}+(\lfloor n/2 \rfloor+1)^a2^{n+1}=O(n^a 2^{n+1})$

• Let $B_n= \sum_{k=1}^n2^k$ for $n\geq 1$ and $B_0=0$

Performing Abel transformation, $$\begin{align}S_{n,a} &=(n+1)^aB_n + \sum_{k=0}^n(k^a-(k+1)^a)B_k\\ &= (n+1)^a(2^{n+1}-2) + \sum_{k=0}^n(k^a-(k+1)^a)(2^{k+1}-2)\end{align}$$

By mean value theorem, depending on the value of $a$, we get the following bounds: $$\left|\sum_{k=0}^n(k^a-(k+1)^a)(2^{k+1}-2)\right|\leq 2|a|\sum_{k=0}^n(k+1)^{a-1}2^{k}$$ $$\left|\sum_{k=0}^n(k^a-(k+1)^a)(2^{k+1}-2)\right|\leq 2|a|\sum_{k=0}^n k^{a-1}2^{k}$$

In both cases, $\sum_{k=0}^n(k^a-(k+1)^a)(2^{k+1}-2)=O(S_{n,a-1})=O(n^{a-1} 2^{n+1})$

As a result, $$S_{n,a} = n^a2^{n+1} +O(n^{a-1} 2^{n+1}) = n^a2^{n+1} + o(n^a2^{n+1})$$

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The same asymptotic estimate holds for integrals, it's essentially the same proof, except less tedious since integration by parts is easier to perform.