Asymptotic behavior of $\sum\limits_{k=1}^n \frac{2^k}{k}$

I'm looking for an asymptotic equivalent of

$$\sum_{0 < k \le n} \frac{2^k}{k}$$

as $n \to \infty$. A plausible candidate seems to be $\frac{2^{n+1}}{n+1}$ (WolframAlpha plot, and the intuitive similarity with $\sum_{k \le n} 2^k = 2^{n+1}$ is also appealing), but my usual tricks seem powerless here.

I've tried:

• Interpreting $x^k/k$ as the primitive of $x^{k-1}$ and setting $x=2$
• Replacing $2^k/k$ with $\int_{x=0}^2 x^{k-1}$
• Reordering the sum's terms to expose log-resembling sub-sums like $\sum_k 1/k$
• Finding lower and upper bounds asymptotically equivalent to $\frac{2^{n+1}}{n+1}$ -- the lower bound is easy ($\sum\limits_{k \le n} \frac{2^k}{k} \ge \sum\limits_{k \le n} \frac{2^k}{n+1} \ge \frac{\sum\limits_{k \le n} 2^k}{n+1} \ge \frac{2^{n+1}}{n+1}$), but the upper bound seems trickier (I couldn't think of a sequence $\varepsilon_k \in o(2^k/k)$ that would make it easy to estimate $\sum_{0 < k \le n} 2^k/k - \varepsilon_k$)
• ... and a few others, to no avail

Any hints?

Solution 1:

For every $n$, $$S_n=\sum_{k=1}^n\frac1k2^k\geqslant\sum_{k=1}^n\frac1n2^k=\frac1n(2^{n+1}-1).$$ On the other hand, for every $u$ in $(0,1)$, $$S_n=\sum_{k\lt un}\frac1k2^k+\sum_{un\leqslant k\leqslant n}\frac1k2^k\leqslant\sum_{k\lt un}2^k+\sum_{un\leqslant k\leqslant n}\frac1{un}2^k\leqslant2^{un+1}+\frac1{un}2^{n+1}.$$ Thus, $$ 2-\frac1{2^n}\leqslant\frac{n}{2^n}S_n\leqslant\frac2u+\frac{2n}{2^{(1-u)n}} $$ which implies $$2\leqslant\liminf_{n\to\infty}\frac{n}{2^n}S_n\leqslant\limsup_{n\to\infty}\frac{n}{2^n}S_n\leqslant\frac2u$$ This holds for every $u\lt1$ hence

$$\lim_{n\to\infty}\frac{n}{2^n}S_n=2.$$

(Which confirms your intuition.)

Likewise, for every real number $\alpha$, $$\lim_{n\to\infty}\frac{n^\alpha}{2^n}\sum_{k=1}^n\frac{2^k}{k^\alpha}=2.$$ Likewise (bis), for every real number $x\gt1$ and every real number $\alpha$,

$$\lim_{n\to\infty}\frac{n^\alpha}{x^n}\sum_{k=1}^n\frac{x^k}{k^\alpha}=\frac{x}{x-1}.$$

Solution 2:

The Euler-Maclaurin summation formula is useful for approximating sums and often reveals the asymptotic behavior with only a few terms. This problem is an interesting application because the precise asymptotic behavior requires summing an infinite number of terms with Bernoulli numbers as coefficients - the terms that are typically neglected.

Using the Euler-Maclaurin summation formula with $f(x) = 2^x/x$

$$\sum_{k=1}^{n}\frac{2^k}{k} = C+\int_{1}^{n}f(x)\,dx + \frac1{2}f(n)+\frac{B_2}{2!}f'(n) + \frac{B_4}{4!}f'''(n)+ \ldots.$$

The integral is the exponential integral which behaves asymptotically as $Ei(x) \sim e^x/x$ as $x \rightarrow \infty:$

$$\int_{1}^{n}f(x)\,dx=\int_{1}^{n}\frac{2^x}{x}\,dx= Ei(n\log2)-Ei(\log2) \sim \frac{e^{n\log 2}}{n\log 2}.$$

Taking the odd-order derivatives we find a pattern

$$f'(x) = \frac{2^x}{x}\left(\log2-\frac1{x}\right)\sim\frac{2^x}{x}\log2\\ f'''(x) = \frac{2^x}{x}\left[\left(\log2-\frac1{x}\right)^3+O(x^{-2})\right]\sim\frac{2^x}{x}(\log 2)^3\\ \ldots$$

Hence,

$$\sum_{k=1}^{n}\frac{2^k}{k} \sim \frac{2^n}{n}\left[\frac1{\log 2}+\frac1{2}+\frac1{\log 2}\left(\frac{B_2}{2!}(\log 2)^2 + \frac{B_4}{4!}(\log 2)^4+ \ldots\right)\right]\,\,(*)$$

The trailing terms can be summed exactly. The generating function for the Bernoulli numbers is $x/(e^x-1)$ where

$$\frac{x}{e^x-1} = \sum_{k=0}^{\infty}\frac{B_kx^k}{k!}.$$

The first two Bernoulli numbers are $B_0 = 1$ and $B_1 = -1/2$. They are zero for odd $n \geq 3$.

Hence,

$$\sum_{k=2}^{\infty}\frac{B_k(\log 2)^k}{k!} = \frac{\log 2}{e^{\log 2}-1}-1 + \frac{\log 2}{2}= \log 2-1 + \frac{\log 2}{2}.$$

Substituting into $(*)$ we get

$$\sum_{k=1}^{n}\frac{2^k}{k} \sim \frac{2^n}{n}\left[\frac1{\log 2}+\frac1{2}+\frac1{\log 2}\left(\log 2-1 + \frac{\log 2}{2}\right)\right]=\frac{2^n}{n}2,$$

and

$$\sum_{k=1}^{n}\frac{2^k}{k} \sim \frac{2^{n+1}}{n}$$