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I don't understand how to show that the following is true:

$$ (1+\frac{\epsilon}{5})^t \leq e^{\frac{t\epsilon}{5}}$$

I know that the exponential function can be defined as:

$$\lim_{n\to \infty} (1+\frac{1}{n})^n = e$$ and then by substitution $n \to nx$ we can obtain that $$\lim_{n\to \infty} (1+\frac{x}{n})^n = \lim_{n\to \infty} (1+\frac{1}{n})^{nx} = e^x$$

Moreover, we also know that the first limit is non-decreasing, thus approaching the value of $e$ from below (thus it should be simple to show the upper bound). My problem is that $\frac{\epsilon}{5}$ does not depend on $t$, so I am not really sure what to do.

Of course, we can do a substitution with $x$ and say $\frac{t\epsilon}{5} = x$ and obtain:

$$(1 + \frac{t\epsilon}{5} \frac{1}{t})^t = e^{\frac{t\epsilon}{5}}$$

but it seems a little bit strange, as we don't change the exponent, nor we have another $x$ on the left hand side of the substitution (i.e. before we had $x \rightarrow nx$, here the $x$ is not appearing on one side).

Is this correct? How can I know that my substitusions are correct? Is there a list of rule / tutorial / book to study? Thanks.

asdf
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2 Answers2

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Write $$\left(1+\frac\epsilon 5\right)^t=\left(1+\frac\epsilon 5\right)^{\frac 5\epsilon \frac\epsilon 5t}=\left[\left(1+\frac\epsilon 5\right)^{\frac 5\epsilon}\right]^{\frac{\epsilon t}5}\le e^{\frac{\epsilon t}5}$$

Andrei
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  • I don't understand. Where is the limit? What goes to infinity here? – asdf Jul 07 '21 at 18:22
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    There is no limit. You just need to show $(1+\epsilon/5)^{5/\epsilon}\le e$ – Andrei Jul 07 '21 at 18:24
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    See https://math.stackexchange.com/questions/167843/show-that-left1-dfrac1n-rightn-is-monotonically-increasing for example – Andrei Jul 07 '21 at 18:35
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If $t$ is an integer, use binomial expansion.

$(1+\frac{x}{n})^n=\sum\limits_{k=0}^n \binom{n}{k}(\frac{x}{n})^k$

$e^x=\sum\limits_{k=0}^\infty \frac{1}{k!}x^k$.

$\binom{n}{k}\frac{1}{n^k}=\frac{n(n-1)(...(n-k+1)}{n^k}{\frac{1}{k!}}\le \frac{1}{k!}$

Net result $(1+\frac{x}{n})^n\le e^x$.

herb steinberg
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