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I came up with $e = \sum_{n=0}^\infty \frac 1 {n!}$ (see here)

I am now trying to prove that this is equivalent to $\lim_{n\to \infty} {(1+\frac1 n)}^n$

In general, how would one go about such a task? Please do not prove what I stated above, I'd like to do that myself! Please tell me what conditions we have to satisfy to show that two expressions to infinity are equivalent?

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Rohil Verma
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    Try expanding $(1+\frac{1}{n})^n$ using the binomial formula. – Tyr Curtis Apr 26 '15 at 16:24
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    The hint above basically tells you the terms of your two sequences are actually the same. What you could do in general is take the difference of your two sequences and show that this converges to zero (assuming that the two sequences have their own respected limits). – Marc Apr 26 '15 at 16:31
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    The only "condition" that can be stated for the case of any two convergent sequences having the same limit, is that the limit of their difference is zero (which is only slightly weaker than just "the limits are the same value"). In this case proving that the difference tends to zero will amount to the same as proving that the second limit is $e$. – GPerez Apr 26 '15 at 16:32
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    See equivalence of the characterizations of the exponential function. – Lucian Apr 26 '15 at 17:30

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