How to evaluate the sum $\sum_{i=1}^n{{n}\choose{i}}\frac{1}{i}$

Question

I'm trying to evaluate the following sum $$ \sum_{i=1}^n{{n}\choose{i}}\frac{1}{i} $$ in order to calculate it (with a computer program, perhaps) for high values of $n$. Is it possible?

Answer 1

It can be written using a hypergeometric function: $$ {\mbox{$_3$F$_2$}(1,1,-n+1;\,2,2;\,-1)}n$$ I don't think there's anything more "closed-form" than that. But the generating function is $$\dfrac{\ln((1-x)/(1-2x))}{1-x}$$

EDIT: Well, a little more closed-form:

$$-{\frac {{2}^{n}\Phi\left( 2,1,n \right) n-{2}^{n}+\Psi \left( n \right) n+1+ \left( i\pi+\gamma \right) n}{n}} $$

where $\Phi$ is the Lerch Phi function and $\Psi$ the digamma function.

Answer 2

$$\sum_{i=1}^{n} \binom{n}{i}\frac{x^i}{i} =\int_{0}^x \frac{(1+t)^n -1}{t}\,dt$$

Letting $u=1+t$:

$$\int_1^{1+x} \frac {u^n-1}{u-1}\,du=\sum_{k=1}^n \frac{(1+x)^k-1}{k}$$

Letting $x=1$, this is:

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

This is unlikely to have a nice closed form without going to special functions.

The nice thing about this formula is that it is easy to inductively compute the values - if $T_n$ is your value then $T_n-T_{n-1}=\frac{2^n-1}{n}$.

Answer 3

For large values of $n$, many terms are such that $i\approx i+1$.

Then

$$\sum_{i=1}^n\binom ni\frac1{i+1}=\frac1{n+1}\sum_{i=1}^n\binom{n+1}{i+1}=\frac{2^{n+1}-n-2}{n+1},$$ which should give a first asymptotic approximation.

A more precise approximation should be possible by developing

$$\frac1i=\frac1{i+1}+\frac a{(i+1)(i+2)}+\frac b{i(i+1)(i+2)}\cdots$$ and expressing the related sums $$\frac{2^{n+k}-P_k(n)}{(n+1)(n+2)\cdots(n+k)}.$$