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I need to find the sum of this series: $1+\frac 12 + \frac 13 + \cdots + \frac 1{50}$

Please help me find the sum of this series.

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There is no nice closed formula for partial harmonic sums. However, there is a technique called partial summation used to estimate these types of sums. Partial summation, or Abel summation, is just integration by parts for the Riemann-Stieltjes integral.

You can look up how to approximate sums by the method of partial summation if you are interested. The estimate it provides for partial harmonic sums is

\begin{align} \sum_{n \leq x}{\frac{1}{n}} &= \log(x) + \gamma + O\left(\frac{1}{x}\right), \end{align}

where $\gamma \approx 0.5772$ is the Euler-Mascheroni constant. Note that the range of the sum is over all positive integers $n$ at most $x$. The $O(1/x)$ term just means the approximation gets better as we chose larger $x$.

For $x = 50$ this formula tells us

\begin{align} \sum_{n \leq 50}{\frac{1}{n}} &\approx 4.4892. \end{align}

If you calculate the sum exactly, the error in the approximation is about $0.00997$.

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It can be shown that your limit corresponds to the harmonic number:

$$H_{50} = \sum^{50}_{n=1} \frac{1}{n},$$

and by using the formula:

$$H_n = \Psi(n+1) - \gamma,$$

with $\Psi$ being the digamma function and $\gamma$ the Euler-Mascheroni constant. Hence:

$$H_{50} = \Psi{(51)} - \gamma \approx 4.4992.$$

Cheers.


To evaluate $\Psi(k)$ use the PolyGamma command in Mathematica and EulerGamma to compute $\gamma$.

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  • $\begingroup$ Could you please enlighten me on the source of the formula in the second equation? Where does it come from? Cheers $\endgroup$ – Abhimanyu Arora Jan 20 '14 at 15:57
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    $\begingroup$ I'm afraid I do not know the exact details but I took it from here a long time ago: mathworld.wolfram.com/HarmonicNumber.html $\endgroup$ – Dmoreno Jan 20 '14 at 16:54
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As is easily verified with your preferred CAS (I used Maple here): $$\sum_{n=1}^{50} \frac1n = \frac{13943237577224054960759}{3099044504245996706400}$$

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$$LCM(1,2,3,4,\ldots,49,50) = 2^5\times 3^3 \times 5^2 \times 7^2 \times \prod_{p\in\mathbb{P},p\le50}p = 2^4\times 3^3\times 5\times 7 \times 47\#$$ $$ = 3\:099\:044\:504\:245\:996\:706\:400, $$ where $47\#$ is primorial number.

Playing with WolframAlpha, one can get exact value: $$ \sum_{k=1}^{50}\dfrac{1}{k} = \dfrac{13\:943\:237\:577\:224\:054\:960\:759}{3\:099\:044\:504\:245\:996\:706\:400} = {\large{4}}\dfrac{1\:547\:059\:560\:240\:068\:135\:159}{3\:099\:044\:504\:245\:996\:706\:400} \approx{4.499205}. $$

Notice it is long and uncomfortable way to find this sum manually for ~30 and more terms of sum.


Perhaps, more comfortable way is:

to find sum of first few terms "manually", and sum of other terms replace with integral:

$$ \sum_{k=1}^{50}\dfrac{1}{k} \approx \sum_{k=1}^{k_0}\dfrac{1}{k} + \int\limits_{k_0+0.5}^{50+0.5} \dfrac{dx}{x} = \sum_{k=1}^{k_0}\dfrac{1}{k} + \ln(50.5)-\ln(k_0+0.5). $$

For $k_0=10$ (for example) we can get $$ \sum_{k=1}^{50}\dfrac{1}{k} \approx \dfrac{7381}{2520} +\ln 50.5-\ln 10.5 \approx\dfrac{7381}{2520} + 1.570598 \approx 4.499\color{gray}{566}. $$

For $k_0=20$ we'll get $$ \sum_{k=1}^{50}\dfrac{1}{k} \approx \dfrac{55\:835\:135}{15\:519\:504} +\ln 50.5-\ln 20.5 \approx 4.4992\color{gray}{88}. $$

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  • $\begingroup$ sir, thanks for your answer but what i did is.. i calculated the harmonic mean of the series by taking the 1st and last term i.e 1 and 1/50 .. so the mean is 2/51. now sum of the series will be = harmonic mean x number of terms. which comes out to be (2/51)x50 = 1.96 aprox. .. what is wrong in my approach ? $\endgroup$ – Rishabh Gour Aug 16 '15 at 9:17
  • $\begingroup$ @Rishabh Gour, consider $4$ terms: $1+\frac12+\frac13+\frac14$. Even if $HM(a_1,a_4)=HM(1,\frac14)=2/5$ and $HM(a_2,a_3)=HM(\frac12,\frac13)=2/5$, it is wrong to suppose that $a_1+a_4 = a_2+a_3$. To apply this trick, arithmetic means must be equal, not harmonic means. $\endgroup$ – Oleg567 Aug 16 '15 at 10:14
  • $\begingroup$ Yeah, now I got it. Thanks a lot :D $\endgroup$ – Rishabh Gour Aug 23 '15 at 6:00

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