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I have the suggestion that the following estimate is true for all $k\geq 2$:

$$\frac{4}{n}\sum\limits_{i=1}^{k-1}\left(\frac{4}{n}+1\right)^i\leq \left(\frac{4}{n}+1\right)^{k+1}+(k+1)$$

I tryed a lot, but couldn't solve it yet. Do you have an idea? If it helps, i can post my approaches.

Thank you!

  • 2
    You might want to come up with a better title.2012-10-15
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    @Asaf: This one looks invitingly like the first line of a poem, though.2012-10-15
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    Isn't this easily solveable using the geometric series? We know that for all $k \ge 1$, $\sum_{i=1}^k q^k = \frac{q^{k+1}-1}{q-1}$...2012-10-15

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$\frac{4}{n}\sum\limits_{i=1}^{k-1}\left(\frac{4}{n}+1\right)^i = \left(1+\frac{4}{n}\right)^k-\left(1+\frac{4}{n}\right) \leq \left(\frac{4}{n}+1\right)^{k+1}$

the left hand of the inequality is a geometrical series.

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    To be more precise, let $q := \frac{4}{n} - 1$. Then $$\frac{4}{n}\sum_{i=1}^{k-1} \left(\frac{4}{n}+1\right)^i = (q-1) \sum_{i=1}^{k-1} q^i = (q-1) \frac{q^k-q}{q-1} = q^k-k = \left(1+\frac{4}{n}\right)^k - \left(1+\frac{4}{n}\right).$$2012-10-15
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    I think you meant $q:= \frac{4}{n} +1$, right?2014-07-04