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?

  • $\begingroup$ I'd think interpreting this as a certain evaluation of a logarithm to be the soundest approach, since one can bring some tools from complex analysis to bear on such. $\endgroup$ – Semiclassical Aug 5 '14 at 17:54
  • $\begingroup$ Call your sum $S_n$. Let $T_n = S_n/2^n$; then it's not hard to show that $T_1 = 1$, $T_n = T_{n-1}/2 + 1/n$. You want to show $T_n \sim 2/(n+1)$. Not sure this helps. $\endgroup$ – Michael Lugo Aug 5 '14 at 18:51
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    $\begingroup$ You mean $1\leq k \leq n$. $\endgroup$ – Mhenni Benghorbal Aug 5 '14 at 19:08

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


(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 $\alpha$,


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    $\begingroup$ Thanks! I find the "This holds for every ... hence" line a bit confusing though :/ Once you multiply both sides by $n/2^n$ and look at $u \to 1$, doesn't the upper bound reduce to $2n+2$ instead of just $2+\varepsilon_n$? Or did I miss something? $\endgroup$ – Clément Aug 5 '14 at 23:01
  • $\begingroup$ Once one multiplies both sides by $n/2^n$ and one considers the limit $n\to\infty$, one sees that the limsup is at most $2/u$. This holds for every $u\lt1$ (no limit here) hence the limsup us $\leqslant2$, QED. $\endgroup$ – Did Aug 6 '14 at 8:23
  • $\begingroup$ Alright, this makes a lot of sense. Very nice solution! $\endgroup$ – Clément Aug 7 '14 at 22:17

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

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


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

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 $n \geq 3$.


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

Substituting into $(*)$ we get

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


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

  • $\begingroup$ Very interesting calculation. I'm accepting the simpler answer, but this is indeed an interesting case where the small trends add up. Nice one! $\endgroup$ – Clément Aug 7 '14 at 22:19

If $u_n=\dfrac{2^n}n$, we have $u_{n+1}-u_n\sim\dfrac{2^n}n=u_n$.

As the serie $\sum u_n$ is nonnegative and diverges, we obtain by Stolz-Cesàro: $$S_n=\displaystyle\sum_{k=1}^n u_k\sim \displaystyle\sum_{k=1}^n (u_{k+1}-u_k)=u_{n+1}-u_1\sim\frac{2^{n+1}}n.$$


Your intuition is correct, we have

$$\frac{a_n}{b_n}=\frac{\sum_{k=1}^n \frac{2^k}{k}}{\frac{2^{n+1}}{n+1}} \to 1$$

indeed by Stolz-Cesaro

$$\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\frac{\frac{2^{n+1}}{n+1}}{\frac{2^{n+2}}{n+2}-\frac{2^{n+1}}{n+1}}=\frac{1}{2\frac{n+1}{n+2}-1}\to 1$$


We note that the average value of


is given by

$$ \frac{1}{n-1} \int_1^n{\frac{2^n}{n} dn} = \frac{1}{n-1} \left[Ei(n\log(2)) - some constant \right] \approx \frac{1}{n-1} Ei(n \log(2))$$

We note that $$e^{x\log(2)}$$ is asymptotically greater than $$Ei(x\log(2))$$ but it appears that I cannot find a constant $c$ between 0 and 1 such that

$$e^{cx\log(2)}$$ is also asymptotically greater suggesting these two asymptotic classes may have a more intricate relationship than is apparent

For all intents and purposes $\frac{2^n}{n}$ is asymptotically equivalent. To find something that matches the difference of terms more closely will require heavier machinery

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    $\begingroup$ Side note: You seem to be using $n$ as both a bound variable under the integral, and unbound outside; that's somewhat confusing. $\endgroup$ – Clément Aug 5 '14 at 20:19
  • $\begingroup$ @Clément I understand intuitively that doing doesn't have any geometric interpretation, but I've never found a time when such a symbolic calculation has ever given me issues. Why is there so much opposition to doing so? It seems like a natural way to incorporate indefinite integration into the frame work of definite integrals $\endgroup$ – frogeyedpeas Aug 5 '14 at 21:26
  • $\begingroup$ Well, you have a definite integral here. It looks a bit like writing $\forall x, \forall x, x-x=0$, which might be formally valid, but is still confusing. $\endgroup$ – Clément Aug 5 '14 at 22:44

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