How to find the sum of the series stated bellow. 1/100 + 1/101 + 1/102 + ......+1/1000
I tried but it doesn't have any patterns to go with.
Please provide the approach for the above
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Jaume Oliver Lafont
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Mary Lopez
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Why do you think you can find it? In particular this sum looks pretty ugly according to WA – kingW3 May 28 '17 at 19:21
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There is no general closed form. All you can find are good approximations. Particularly, you are looking for $H_{1000}-H_{99}$ – infinitylord May 28 '17 at 19:26
3 Answers
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As in Summing Finitely Many Terms of Harmonic Series: $\sum_{k=a}^{b} \frac{1}{k}$, for an approximated result, we may use $$ \sum_{k=a}^b \frac{1}{k} \approx \log\left(\frac{2b+1}{2a-1}\right) $$
to obtain
$$\sum_{k=100}^{1000} \frac{1}{k}\approx \log\left(\frac{2001}{199}\right) \approx 2.308097...$$
while first digits of the exact result are $2.308093...$
Jaume Oliver Lafont
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1After looking at your different answers to this type of problems, I found an interesting paper(link given in my answer). Cheers. – Claude Leibovici May 29 '17 at 06:42
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Hoping that this could be of interest to you.
I found this paper very interesting since it shows several approximations of the harmonic numbers. In particular $$H_n\approx \log \left(n+\frac{1}{2}\right)+\gamma+\frac{5}{30 \left(2n+1\right)^2+{21}}$$ which is said to overestimate the result by $$\frac{2071}{12600 (2 n+1)^6}$$ This makes $$\sum_{i=a}^b\frac 1i=H_b-H_{a-1}\approx \log \left(\frac{2 b+1}{2 a-1}\right)+\frac{5}{30 (2 b+1)^2+21}-\frac{5}{30 (2 a-1)^2+21}$$ Using $b=1000$ and $a=100$ this leads to $$\sum_{k=100}^{1000} \frac{1}{k}\approx\log \left(\frac{2001}{199}\right)-\frac{198220000}{47569582236867}\approx \color{red} {2.30809334291072}73$$ while the "exact" value would be $\approx \color{red} {2.3080933429107246}$
Computing the overestimate from the given formula, it is $\approx 2.65 \times 10^{-15}$ which matches the above values.
Claude Leibovici
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@JaumeOliverLafont. Thanks. In fact, when I saw, in aprevious answer of yours on this topic, I just wondered if there was anything new in the area of the asymptotics of harmonic numbers and, by pure luck, I found this paper. What I think to be very interesting are the upper and lower bounds provided for the one in my answer. – Claude Leibovici May 29 '17 at 08:44
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Just to be sure, do you mean formula (1.14) in Theorem (1.7)? – Jaume Oliver Lafont May 29 '17 at 08:51
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Maybe this correction that improves accuracy can be related to the question https://math.stackexchange.com/questions/2301252/an-integral-and-series-for-log5-frac85 – Jaume Oliver Lafont May 29 '17 at 11:29
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The approximated value of the "Harmonic Number": $$ H_n=\sum_{i=1}^n\frac{1}{i} \approx ln(n)+\gamma+\frac{1}{2n}\\ \gamma=0.577 $$ The exact value of the requested sum: $$ H_{1000}-H_{99}=2.30809334 $$
Brethlosze
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