Proof that: ∑n=0∞1n!=e

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Proof that:  ∑n=0∞1n!=e

sainareon2 · Accepted Answer

By definition,e=limn→∞(1+1n)nUsing the binomial theorem, the k-th term of the development is(nk)1nk=n(n−1)(n−2)…(n−k+1)k!⋅n⋅n…nandlimn→∞(nk)1nk=1k!For example,(1+11000)1000=10!+11!+0.992!+0.9970023!+0.9940109944!…&#039;

trasahed · Accepted Answer

Since:(xnn!)′=xn−1(n−1)!=xmm!with m=n-1You may know that, and start with, a fundamental property of the natural exponential: it is equal to its derivative. So, suppose thatf(x)=∑n=0∞anxnis equal to its derivative. Then formally (I am skipping issues on convergence),f′(x)=∑n=1∞nanxn−1thus by reindexing:f′(x)=∑n=0∞(n+1)an+1xnthen, one should have, term by term:an=(n+1)an+1hencean+1=ann+1=a0(n+1)!Since the exponential is the reciprocal to the log⁡, you require that f(0)=1, hence a0=1. So naturallyf(x)=∑n=0∞xnn!=exandf(1)=∑n=0∞1n!=e

RizerMix · Accepted Answer

Here is one approach.Let b&amp;gt;1. If you compute the derivative of the function bx, you find that the answer is just bx multiplied by an (annoying) constant.There is a value of b for which this constant is equal to 1. That&#039;s nice! With this special value of b, the derivative of bx is just bx, the same thing we started with. That&#039;s a very neat property for a function to have.This special value of b is e=2.718...It is now easy to compute the Taylor series of the function ex (centered at 0). We find thatex=1+x+x22!+x33!+... (1)This comes directly from the Taylor series formulaf(x)=f(x0)+f′(x0)(x−x0)+f″(x0)2!(x−x0)2+...Plugging x=1 into (1) yieldse=∑n=0∞1n!

Proof that: \sum_{n=0}^\infty\frac{1}{n!}=e

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