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Is inequality $$\Big (1+\frac{1}{k(k+2)}\Big )^k \leq 1+\frac{1}{k+1}$$ is true for all $k \in \mathbb{N}$ ? In my previous question the $k+2$ in the LHS of the inequality was $k+1$. If it was the problem, then it can be settled using binomial theorem. In this case, Binomial theorem is not anymore applicable.what approach could be used here. thanks..

Bart Michels
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Keneth Adrian
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    This is not a duplicate guys, the denominator is different. It is $k+2$ in this, and $k+1$ in the other. Please look carefully before closing. – Calvin Lin Sep 07 '13 at 17:55
  • @HagenvonEitzen: I have asked this question with $k+1$. But my bad it should have been $k+2$. This question.. according to Mercy, there is a significant difference in the changes I did..thanks – Keneth Adrian Sep 07 '13 at 18:03

1 Answers1

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Show that

$$\left(1 + \frac{1}{k(k+2)} \right) ^{k(k+2)} < \sum_{i=0}^{\infty} \frac{1}{i!} < \left( 1 + \frac{1}{k+1}\right)^{k+2}.$$

This can be done by Binomial Theorem expansion.

The motivation for this comes from knowing that $ \lim_{x\rightarrow \infty} (1 + \frac{1}{x} )^x $ tends towards $e$ from below, and $ \lim_{x\rightarrow \infty} ( 1 + \frac{1}{x} )^{x+1} $ tends towards $e$ from above.

Hint: Raise both sides to the $k+2$ power, and apply the above.

$(1 + \frac{1}{k(k+2)} ) ^{k(k+2)} < e < ( 1 + \frac{1}{k+1})^{k+2}$

In your previous question, you tagged it with calculus, so I'm assuming that the first statement is a valid assumption. In fact, we can show that $ ( 1 + \frac{1}{x})^x < e < ( 1 + \frac{1}{x} )^{x+1}$ by using the Binomial Theorem and the power series expansion of $e$.

Calvin Lin
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  • Without using limits, is there another way of proving this? – Keneth Adrian Sep 07 '13 at 18:01
  • @KenethAdrian Just expand the powered inequality using the binomial theorem. The first two terms are equal, and the rest should easily compare, since they already easily compare to $e$. – Calvin Lin Sep 07 '13 at 18:02
  • Oh I see, that was a clever approach.. thank you. – Keneth Adrian Sep 07 '13 at 18:07
  • @KenethAdrian Yea, it helps to be aware of different ways of showing the same fact, which gives you many different approaches to the question. The motivation for the hint is very strongly motivated by the fact, which is why I presented that version. – Calvin Lin Sep 07 '13 at 18:09
  • the power series expansion of $e$ is an infinite sum and using the binomial coefficient expansion of the $(1+ \frac{1}{x})^{x+1})$ is finite.. how can we be sure that this infinite sum on the LHS is lesser than tha finite sum on the right side?? thanks – Keneth Adrian Sep 07 '13 at 18:17
  • @KenethAdrian Think about it through Calculus, we know that we can bound the Taylor Polynomial using the remainder. This converts the infinite series to a finite series. For example, $ e^x < 1 + x + \frac{e^x}{2!} x^2$, which shows us that $ e < 2 + \frac{e}{2} < (\frac{3}{2})^3$, which does it for $k = 1$. – Calvin Lin Sep 07 '13 at 18:27