GATE IT 2004 | Question: 34

Question

Let $H_1, H_2, H_3,$ ... be harmonic numbers. Then, for $n \in Z^+$,  $\sum_{j=1}^{n} H_j$ can be expressed as

• A. $nH_{n+1} - (n + 1)$
• B. $(n + 1)H_n - n$
• C. $nH_n - n$
• D. $(n + 1) H_{n+1} - (n + 1)$

Comments

present in current syllabus?

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I thought $H_j$ is n-th terms of harmonic series, been trying to find closed form for $H_j$ for past 1 hour 💀

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i assumed $H_n=\frac{1}{n}$, which is wrong. 

$H_n$ represents the sum of reciprocal of first $n$ natural numbers. 

Wikipedia Link

Answer 1

The $n^{th}$ Harmonic Number  is defined as the summation of the reciprocals of all numbers from $1$ to $n$.

$$H_n = \sum_{i = 1}^n \frac1 i = \frac1 1 + \frac1 2 + \frac1 3 + \frac1 4 + \dots + \frac1 n$$

Lets call the value of $\sum_{j = 1}^n H_j$ as $S_n$

Then,

$\begin{align}
S_n &= H_1 + H_2 + H_3 + \dots + H_n\\[1em]
&= \small \overbrace{\left ( \color{red}{\frac1 1} \right )}^{H_1}
 + \underbrace{\left (\color{red}{\frac1 1} + \color{blue}{\frac1 2} \right )}_{H_2}
 + \overbrace{\left (\color{red}{\frac1 1} + \color{blue}{\frac1 2} + \color{green}{\frac1 3} \right )}^{H_3}
 + \dots
 + \underbrace{\left (\color{red}{\frac1 1} + \color{blue}{\frac1 2} + \color{green}{\frac1 3} + \dots + \frac1 n \right )}_{H_n}\\[1em]
&=\small  \color{red}{n \times\frac1 1}+ \color{blue}{ (n-1) \times\frac1 2} + \color{green}{(n-2) \times \frac1 3} + \dots + 1 \times \frac1 n\\[1em]
&= \sum_{i = 1}^n \left (n - i + 1 \right ) \times \frac1 i\\[1em]
&= \sum_{i = 1}^n \left ( \frac{n + 1}{i} - 1 \right )\\[1em]
&= \left ( \sum_{i = 1}^n \frac{\color{red}{n+1}}{i}\right ) - \color{blue}{\left ( \sum_{i = 1}^n 1\right )}\\[1em]
&= \left (\color{red}{(n+1)} \times \underbrace{\sum_{i = 1}^n \frac1 i}_{=H_n}\;\right ) - \color{blue}{n}\\[3em]
\hline
\large S_n &= \large (n+1)\cdot H_n - n
\end{align}$

Hence, the answer is option (B).

Comments

Here Option B & D are exactly same.

Option D = (n + 1) Hn+1 - (n + 1)                                --- Equation 1

and we know Hn+1 = Hn + 1/n+1, lets put it in equation 1

Option D = (n + 1)(Hn + 1/(n+1)) - (n+1)

               = (n + 1)Hn + 1 - n - 1

               = (n + 1)Hn - n

               = Option B

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Yes , B&D are exactly same.

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excellant explanation but it takes time to understand the steps, and needs good practice to do it in exam , Thank you @Pragy Agarwal & @rahul sharma 5 for ur wonderful proofs