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I have a question which might be too simple for this site but I really tried many ideas without coming to a solution. This is assignment from elementary school in which I am trying to help and the solution should be relatively simple but somehow I cannot figure out the correct approach. The assignment is as follows i.e. calculate the sum: $$\frac{1}{1+2} + \frac{1}{2+3} + \frac{1}{3+4} + … + \frac{1}{98+99}+\frac{1}{99+100}$$

abkds
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user120250
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    At least one can do the additions in the denominators to get $$\frac13+\frac15+\frac17+\frac19+\cdots+\frac1{197}+\frac1{199} $$ But I don't see any obvious way to go from there. If it had been $$\frac1{1+2}+\frac1{1+2+3}+\frac1{1+2+3+4}+\cdots $$ or $$\frac1{1\times 2}+\frac1{2\times 3}+\frac1{3\times 4}+\cdots $$ instead there would have been room for some telescoping tricks, but ... – hmakholm left over Monica Feb 17 '14 at 12:36
  • Yes there must be some trick, and I really checked if I read the assignment correctly, but it seems that this is the case, and I am simply out of ideas... – user120250 Feb 17 '14 at 12:40
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    Using Mathematica you get a big fraction which is pretty much useless. Are you sure the question didn't contain anything else? because I have seen similar with $$\frac{1}{\sqrt1+\sqrt2}+\frac{1}{\sqrt2+\sqrt3}...$$That one actually has a nice answer – Ali Caglayan Feb 17 '14 at 13:01
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    Because your sum is 2.28931731367974379839114596726193406285803856128546999419511292662827 3526387943042699856735813198916 – Ali Caglayan Feb 17 '14 at 13:04
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    Is this really an assignment from elementary school ? As Alizter commented, any CAS provides a fraction which is just a monster $\frac{6019484180875094293775771025953918105702575294840191045657968324602140467851923 044049097}{26351061627572364424958263030846984955655811155090408924128673587283907 66099042109898375}$. – Claude Leibovici Feb 17 '14 at 13:06
  • @Alizter. I tried your with square roots but what I got is just awful ! What did you do to make it nice ? – Claude Leibovici Feb 17 '14 at 13:10
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    @ClaudeLeibovici Here – Ali Caglayan Feb 17 '14 at 13:11
  • Thank you for effort guys. Yes this is elementary school. The only thing which might be is that they did some mistake when typing it. I really cannot offer better explanation. – user120250 Feb 17 '14 at 13:17
  • @Alizter. So simple and so beautiful !m Thanks for the link. Cheers. – Claude Leibovici Feb 17 '14 at 13:17
  • this is gonna look like $$ \sum_{n=0}^{100} \frac{1}{3+2n} $$ which comes out to about $2.294243422054128034844347937705283816552619842073647334096590...$ not exactly a clean number... – Ephraim May 07 '14 at 14:41

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I think there isn't a nice closed form for this sum. But if we let $H_n$ denote $1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}$, then

$\frac{1}{1 + 2} + \frac{1}{2 + 3} + ... + \frac{1}{99 + 100} = \frac{1}{3} + \frac{1}{5} + ... + \frac{1}{199} = \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{200} - 1 - (\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + ... + \frac{1}{200}) = H_{200} - 1 - \frac{1}{2}(1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{100}) = H_{200} - \frac{1}{2}H_{100} - 1$

But the harmonic sum does not have a nice closed form as far as I knew.

zscoder
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