# Is the summation $\sum_{i=1}^{n}\frac1{i} \binom{n}{i}$ possible?

I want to compute the following sum: $$\sum\limits_{i=1}^{n} \frac{{n\choose{i}}}{i}$$

What I have done so far:

We know that $$(1+x)^n=\sum\limits_{r=0}^{n} {n\choose{r}}x^r$$ so, $$\frac{(1+x)^n-1}{x}=\sum\limits_{i=1}^{n} {{n\choose{i}}}x^{i-1}$$ therefore, upon integration we get, $$\int\limits_{0}^{1}\frac{(1+x)^n-1}{x}dx=\sum\limits_{i=1}^{n} \frac{{{n\choose{i}}}}{i}$$

I cannot get any further with the LHS of the above equation.

1. Is such an integration possible (why so)?
2. Are there any other approximations for the sum?
• The integral admits different summation formula: $$\int_{0}^{1}\frac{(x+1)^n-1}{x}\,\mathrm{d}x=\int_{0}^{1}\left(\sum_{k=1}^{n}(1+x)^{k-1}\right)\,\mathrm{d}x=\sum_{k=1}^{n}\frac{2^k-1}{k}.$$ Commented Aug 4, 2020 at 20:39
• $\displaystyle\int_{0}^{1}{\left(\, x + 1\, \right)^{n} - 1 \over x}\,\mathrm{d}x = n\ \mbox{}_3\mathrm{F}_{2}\left(\, 1,1,1 - n;2,2;-1\,\right)$. See this link. Commented Aug 5, 2020 at 3:45

To expand a little on @Sangchul Lee's comment - after the substitution of $${y=1+x}$$ the integral becomes

$${\Rightarrow \int_{1}^{2}\frac{y^n-1}{y-1}dy}$$

(This substitution isn't "necessary" for the next part, however it makes it a bit clearer). Now we can use a special factoring formula:

$${a^n-b^n = (a-b)(a^{n-1} + a^{n-2}b + ... + ab^{n-2} + b^{n-1})}$$

To get

$${\int_{1}^{2}\frac{(y-1)(y^{n-1} + y^{n-2} + ... + y + 1)}{(y-1)}dy=\int_{1}^{2}y^{n-1} + y^{n-2} + ... + y + 1dy}$$

Evaluating that last integral gives us the sum

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

So overall you have that

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

Other than that though, I'm not sure if there's any more useful form. WolframAlpha gives some very nasty looking closed forms involving special functions.

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ \begin{align} \sum_{i = 1}^{n}{{n \choose i} \over i} & = \sum_{i = 1}^{n}{n \choose i}\int_{0}^{1}t^{i - 1}\,\dd t = \int_{0}^{1}\sum_{i = 1}^{n}{n \choose i}t^{i}\,{\dd t \over t} = \int_{0}^{1}\bracks{\pars{1 + t}^{n} - 1}\,{\dd t \over t} \\[5mm] & = \int_{1}^{2}{1 - t^{n} \over 1 - t}\,\dd t = \int_{0}^{2}{1 - t^{n} \over 1 - t}\,\dd t\ -\ \underbrace{\int_{0}^{1}{1 - t^{n} \over 1 - t}\,\dd t}_{\ds{H_{n}}} \\[5mm] & = \int_{0}^{2}{1 - t^{n} \over 1 - t}\,\dd t - H_{n} \end{align}