Prove that $\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\cdots-\frac{1}{2009}+\frac{1}{2010}<\frac{3}{8}$

Oh my, I feel embarrased for not knowing how to solve such an elementary problem but I'm really stuck on this one. I mean - I tried grouping them (I mean - these pairs with minus in-between) to show that the first pair gives $\frac{1}{6}$ and from there it's descending so it has to be less than $\frac{3}{8}$ but it doesn't seem to be a good idea after all. So I tried to remodel the pair-thinking and noted that: $$\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}$$ which leads to pairing the initial sequence to the form of: $$\frac{1}{2\cdot3}+\frac{1}{4\cdot5}+\frac{1}{6\cdot7}+\cdots+\frac{1}{2008\cdot2009}+\frac{1}{2010}$$ but it still seems to be leading nowhere. How should I approach it?


You may be familiar with Leibniz' result on an alternating monotonically decreasing (in absolute value) series: the partial sum is above (resp. below) the sum, if the last included term was positive (resp. negative). That gives us a clue. Compute $$ \frac12-\frac13+\frac14-\frac15+\frac16-\frac17+\frac18=\frac{307}{840}\approx0.365<3/8. $$ If you have not heard of Leibniz, you may simply observe that $-1/9+1/10<0$, $-1/11+1/12<0$, $-1/13+1/14<0$, $\ldots$

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    $\begingroup$ Unfortunately, I haven't heard about Leibniz. But regarding the last part of your post - OK, I get the pairing you used to show that we have simply $\frac{1}{2}$+some_negative_numbers but how do we prove that those negative numbers sum to a number big enough so that when it's added to 1/2, we still get something less than 3/8? $\endgroup$ – Straightfw Apr 22 '12 at 18:30
  • $\begingroup$ @Straightfw, that's what the first part seeks to remedy. We need to do a little bit of dirty work, and check that the sum up to $+1/8$ is below the bound $3/8$. With that task out of the way, the rest follows nicely. $\endgroup$ – Jyrki Lahtonen Apr 22 '12 at 18:33
  • $\begingroup$ Oh, OK, I see - thought it was something to use if I decided to go with Leibniz :) Nice, thank you very much! $\endgroup$ – Straightfw Apr 22 '12 at 18:39

For a similar approach to Jyrki's, we use the result that the alternating harmonic series converges to $\ln 2$. Let's subtract one from both sides of your inequality, then change all the signs to get $$1 - \dfrac{1}{2} +... + \dfrac{1}{2009} - \dfrac{1}{2010} > \dfrac{5}{8}$$and denote the left hand side of this new inequality by $(\#)$. Then $$ (\#) + \sum_{n = 2011}^{\infty} \dfrac{(-1)^{n+1}}{n} = \ln 2 \approx 0.6931$$

So $$( \#) \approx 0.6931 - \sum_{n=2011}^{\infty}\dfrac{(-1)^{n+1}}{n}$$

Which is greater than $\dfrac{5}{8} = 0.625$ if we can bound the tail of this infinite sum (as in the comments to Jyrki's answer). I'll leave that as an exercise :)


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