Is this true? $$\left(1+\frac{1}{n}\right)^n=1+1/1!+1/2!+1/3!+1/4!+1/5!+\cdots $$($n$ times) or ($n+1$ times)?
If yes how to prove it and were there any proof of it?
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4No, this is not true - just try evaluating both sides for $n=3$ – Old John Dec 18 '13 at 20:58
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3Did you try to actually evaluate it for some small $n$? – Listing Dec 18 '13 at 20:58
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3That two sequences have the same limit doesn't mean they're equal – M Turgeon Dec 18 '13 at 20:59
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1in fact i made a mistake, i missed a one. – user123456789 Dec 18 '13 at 20:59
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i corrected it now – user123456789 Dec 18 '13 at 21:00
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1Still wrong. The dots on RHS suggest a limit (it's the number e=2.718..., irrational number), while on the LHS there is a rational number, function of $n$. – Jean-Claude Arbaut Dec 18 '13 at 21:02
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1Even for $n=2$, the left side is $(1+1/2)^2=\frac 94$, while the right is $1+1/1!+1/2!=\frac {10}4$ – Ross Millikan Dec 18 '13 at 21:02
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[(1+1/n)^n where n is the number you think about] of thanks – user123456789 Dec 18 '13 at 21:04
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Coincidentally, I was wondering the same thing a little while back. You might be tempted to think it's true, but it isn't. However, it is true that the terms become closer and closer as n grows larger and larger. But they are never equal. – MPW Dec 18 '13 at 21:05
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I think about nothing, thanks. If you want to write a limit, write it (or if you don't know how to write a limit in latex, ask us to write it). Here it's just mathematical rubbish. – Jean-Claude Arbaut Dec 18 '13 at 21:05
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@Ross sorry I didn't recognize your comment... – draks ... Dec 18 '13 at 21:08
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@user123456789 : your title does not match your question. Your title has a finite sum, and the right-hand side of the equation in your question is an infinite sum, whether this is what you intend or not. You should fix your question for the sake of future readers of this question. – Stefan Smith Dec 18 '13 at 21:18
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@draks...: I evaluated each side of the equation out to $n=2$ and found they did not agree. – Ross Millikan Dec 18 '13 at 21:49
3 Answers
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No, it not true.
Apply binomial theorem: $$ \left(1+\frac{1}{n}\right)^n = \sum_{m=0}^n \binom{n}{m} \frac{1}{n^m} = \sum_{m=0}^n \frac{n (n-1) \cdots (n-m+1)}{n^m} \frac{1}{m!} = \sum_{m=0}^n \frac{1}{m!} \prod_{k=0}^{m-1} \left(1-\frac{k}{n}\right) $$
Sasha
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While the way you posed the question is not true, a slight modification yields something like you want. The LHS is a well known limit and the RHS is the Taylor polynomial (evaluated at 1). $$ \lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^n=e^1=\sum_{k=0}^\infty\frac{x^k}{k!}\bigg|_{x=1} $$ As others have pointed out, these two sequences have the same limit, but there is no guarantee they are equal term for term.
Laars Helenius
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7The original question did not have the limit applied. They are not equal except in the limit. – Ross Millikan Dec 18 '13 at 21:01
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2I agree. I should answer the question as written, not as I think it should have been written, right! ;) – Laars Helenius Dec 18 '13 at 21:02
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OP, any undergraduate text should provide the details you need. – Laars Helenius Dec 18 '13 at 21:05
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No and evaluating $n=2$ would have been enough to see this: $$ \left(1+\frac12\right)^2=1+1+\frac14=2.25 \neq 2.5=1+\frac1{1!}+\frac1{2!} $$
draks ...
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I personally prefer a simple, explicit counterexample like the one you gave. – Stefan Smith Dec 18 '13 at 21:25