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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

  1. $nH_{n+1} - (n + 1)$
  2. $(n + 1)H_n - n$
  3. $nH_n - n$
  4. $(n + 1) H_{n+1} - (n + 1)$
in Combinatory by Boss (16.3k points)
edited by | 1.6k views
+5
\begin{array}{c c c}
H_{1} &= &\frac{1}{1} &&&&&(+)\\
H_{2} &= &\frac{1}{1} &+ &\frac{1}{2}&&&(+) \\
H_{3} &= &\frac{1}{1} &+ &\frac{1}{2} &+ &\frac{1}{3}&\\
\hline
 4*H_3&=& \frac{1}{1} &+ &\frac{1}{2} &+ &\frac{1}{3}&(+)\\
&& \frac{1}{1} &+ &\frac{1}{2} &+ &\color{red}{\frac{1}{3}}&(+)\\
 &&\frac{1}{1} &+ &\color{red}{\frac{1}{2}} &+ &\color{red}{\frac{1}{3}}&(+)\\
&&\color{red}{\frac{1}{1}} &+ &\color{red}{\frac{1}{2}} &+ &\color{red}{\frac{1}{3}}&\\
\hline
(3+1)*H_3&=& \color{red}{1}+\frac{1}{1}*3 &+&\color{red}{1}+\frac{1}{2}*2&+&\color{red}{1}+\frac{1}{3}*1 &= S_3 +\color{red}{3}\\
\hline
\end{array}
So, $S_n= (n+1)H_n - n$
0
wonderful Question !
0
present in current syllabus?

2 Answers

+69 votes
Best answer

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).

by Boss (22.8k points)
edited by
+7
Take n =2 and then find first two harmonic numbers.

And then find their sum.And then use options/only b will satisfy:)
+1
Is both B) and D) are correct answers ...?
0
With this method, none of the options are matching....... I'm getting 5/2 for both B and D, instead of 3/2, when n=2
+1
For such questions, it's fast to check using specific values.

Consider n=4

H1=1

H2=3/2

H3=11/6

H4=25/12

H(n+1) i.e. H5=137/15

Summation of H1 till H4 is 77/12 which is only given by option (b).

Hence, the answer.
0

Ayush Upadhyaya 

G5= 137/60 and for n=4,77/12 both b and d will satisfy

n=2,3,4,5 i tried and both b and d are satisfied.I dont think we can use this approach.Earlier i used this approach but i approximated 1/3 in calc but if i dont than both b and d are giving same answer always from n=2 to 6.

+6

In the continuation to above proof, from last line of the proof in he given answer to show that it is equal to D also.SO bith B and D are true here.

0
nice explanation @ pragy agrawal sir
0
That is right you have to take sum of first  two terms not the term itself
0
I guess both B & D are correct. Please update the answer as B & D.
0

$D$ also true for every case.

Example:

$H_{1}+H_{2}+H_{3}+H_{4}+H_{5}=1+\dfrac{3}{2}+\dfrac{11}{6}+\dfrac{25}{12}+\dfrac{137}{60}=\dfrac{522}{60}=\dfrac{87}{10}$

$H_{6}=\dfrac{147}{60}$

$D)\ 6*H_{6}-6=6*\dfrac{147}{60}-6=\dfrac{87}{10}$


Can try for other values also. If got something which is proving D as wrong. Do tell.

–4 votes

The correct answer is, (B) (n + 1)Hn - n

by Loyal (8k points)
Answer:

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