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The question:

Using $S_n = \sum_{k=1}^{n}H_k$ where $H_k$ are the harmonic numbers, show $S_n = (n+1)H_n - n$.

So far I have

$S_n = \sum_{k=1}^{n} H_k = \sum_{k=1}^{n} \sum_{j=1}^{k}\frac{1}{j} $

is there perhaps some way to change the summation index?

Or would the next step come from

$S_n = \sum_{k=1}^{n}[1+ \frac{1}{2} + \frac{1}{3} + ...+\frac{1}{k-1} + \frac{1}{k}]$

Any help would be appreciated.

Martin Sleziak
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Pablo
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3 Answers3

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$$ \begin{align} \sum_{k=1}^nH_k &=\sum_{k=1}^n\sum_{j=1}^k\frac1j\tag{1}\\ &=\sum_{j=1}^n\sum_{k=j}^n\frac1j\tag{2}\\ &=\sum_{j=1}^n\frac{n-j+1}{j}\tag{3}\\ &=(n+1)\sum_{j=1}^n\frac1j-\sum_{j=1}^n1\tag{4}\\ &=(n+1)H_n-n\tag{5} \end{align} $$ Explanation:
$(1)$: expand $H_k$
$(2)$: exchange order of summation (we are summing over all $1\le j\le k\le n$)
$(3)$: summing $n-j+1$ identical copies of $\frac1j$
$(4)$: pull out the $n+1$ from the $\frac{n+1}{j}$ and separate the $\frac jj$
$(5)$: simplify

robjohn
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A change of the order of summation gives the result: $$S_n=\sum_{j\leqslant k\leqslant n}\frac 1j=\sum_{j=1}^n\sum_{k=j}^n\frac 1j=\sum_{j=1}^n\frac{n-j+1}j=(n+1)H_n-n.$$

Davide Giraudo
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  • Could you please show how you change the order of summation? – Pablo Mar 19 '14 at 21:43
  • Write $[P]=1$ if $P$ is true, and $[P]=0$ if not. Then $S_n=\sum_{j,k=1}^nj^{-1}[j\leq k]=\sum_{k=1}^n\sum_{j=1}^nj^{-1}[j\leq k]=\sum_{j=1}^n\sum_{k=1}^nj^{-1}[j\leq k]$. – Davide Giraudo Mar 19 '14 at 21:50
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Hint: Use induction on $n$.

The base case is $n=1$, which is obvious: $S_1=H_1=1$, while $(1+1)H_1-1=2-1=1$.

Can you see how to relate $S_{n+1}$ to $S_n$ for the inductive step?

Nick Peterson
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