e= appears to be defined as the sum of the series 1/n! as n goes from zero to infinity.  But this implies that the first term is 1/0! which appears to violate the rule against division by zero
54
How can I prove  from first principles that $0!$ is equal to $1$?
56578
$$ \sum_{i=0}^\infty \frac{(-1)^n}{n} $$  This alternating series fails the p-series test because the exponent of n = 1.  Yet it seems to pass the alternating series test.   1 - $a_n$ must be positive. True.  2 - Terms must be decreasing. $\frac{d}{dn} 1/n = -n^{-2}$, which is &lt; 1. True.  3 - $ \lim_{n\rightarrow\infty} 1/n = 0  $ True.  $(-1)^n/n$ is clearly a divergent series, so why does it pass the AST?
34858
Does the following series have a closed-form expression:  $$\sum_{k=0}^{\infty} \frac{z^k}{(k!)^2}$$  I know that it must converge because:  $$\sum_{k=0}^{\infty} \frac{z^k}{k!} = e^z$$  and the $(k!)^2$ denominator obviously increases more quickly than the $k!$ denominator.  This problem came up in computing the probability of a draw in a football match with each team's goal scoring modeled as a Poisson process.  Thanks,  John
117288
Consider the series expansion for the exponential function at x = 1:  $$a_n := \sum\limits_{i = 0}^n{\frac{1}{i!}}$$  I want to prove that this is a Cauchy sequence, using the remainder formula for the exponential function. (That's the term for it I'm familiar with, anyway.) It would be nice if you could comment on whether this proof is correct or if it could be simplified without using additional theorems or such.  For the Cauchy criteria, for each $\epsilon &gt; 0$ there must be a $N(\epsilon) \in N$ so that for all $m, n &gt; N(\epsilon)$, $ |a_m - a_n| &lt; \epsilon$.  So, take any $\epsilon &gt; 0$. Now, choose $N(\epsilon)$ so that:  $${\frac{1}{(N+1)!}} &lt; \frac{\epsilon}{2}.$$  (This is always possible because ${\frac{1}{n!}}$ goes to 0 when $n$ goes to infinity.)  Now, according to the partial sum formula, we got that:  $$\sum\limits_{i = 0}^{\infty}{\frac{1}{i!}} - \sum\limits_{i = 0}^{N(\epsilon)}{\frac{1}{i!}} = \sum\limits_{i = N(\epsilon)}^{\infty}{\frac{1}{i!}} &lt; \epsilon$$  So if we choose any $m, n &gt; N(\epsilon)$, and assume that $m &gt; n$ (because when $m = n$, the Cauchy criteria is obviously true), it follows that:  $$ \sum\limits_{i = 0}^{m}{\frac{1}{i!}} - \sum\limits_{i = 0}^{n}{\frac{1}{i!}} = \sum\limits_{i = n}^{m}{\frac{1}{i!}} &lt; \sum\limits_{i = N(\epsilon)}^{\infty}{\frac{1}{i!}} &lt; \epsilon$$  which proofs the statement.  Edit: the remainder formula I'm familiar with and that I've been using states that:  $$\sum\limits_{i = 0}^{\infty}{\frac{1}{i!}} - \sum\limits_{i = 0}^{n}{\frac{1}{i!}} &lt; 2 \frac{1}{(n+1)!} $$  if I recall correctly. Using this, can't we conclude that $\sum\limits_{i = N(\epsilon)}^{\infty}{\frac{1}{i!}} &lt; \epsilon $ ?
82381
Is the following series convergent?  $$\sum_{n=1}^{\infty}\frac{e^n\,n!}{n^n}$$  I treid the Ratio and Root tests, but both of them failed.
46652
Why does the series $\sum_{n=1}^∞ \ln ({n \over n+1})$ diverges? I'm looking for an answer using the comparison test, I'm just not sure what I can compare it to.   And can I have some tips on what to look at when handling with series that have logarithms in the expression?   Thanks in advance!
49817
Lets define the following series $$S_n =\sum_{i=1}^{\infty} \frac 1 {i(i+1)(i+2)...(i+n)}.$$ I know that $S_0$ does not converge so let's suppose $n \in N$ and define $S_n$ for some $n$. We have $S_1=1$ , $S_2=\frac1 4$ , $S_3=\frac 1 {18}$ , $S_4=\frac 1{96}$ , $S_5=\frac 1{600}$ etc.. the numerator in all results is 1the pattern in denominator is $[1,4,1,96,600...]$ and   that is equal to $n*n!$ Finally I want to prove the general equality: $\sum_{i=1}^{\infty} \frac 1 {i(i+1)(i+2)...(i+n)}=\frac 1 {n*n!}$
6241
What is the value of the following sum?  $$\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \cdots$$  The possible answers are:  A. $e$ B. $\frac{e}{2}$ C. $\frac{3e}{2}$ D.  $1 + \frac{e}{2}$  I tried to expand the options using the series representation of $e$ and putting in $x=1$, but I couldn't get back the original series. Any ideas?
31010
I tried to evaluate $$\lim_{x\to0}\frac{(1+x)^{1/x}-e}x$$but all my efforts were in vain . L'Hospital rule becomes very complicated.
94341
How to show that $\sum\limits_{n = 0}^{\infty} \frac {1}{n!} = e$ where $e = \lim\limits_{n\to\infty} \left({1 + \frac 1 n}\right)^n$?   I'm guessing  this can be done using the Squeeze Theorem by applying the AM-GM inequality. But I can only get the lower bound.   If $S_n$ is the $n$th partial sum of our series then,   $$ \left({1 + \frac 1 n}\right)^n = 1 + n\cdot\frac{1}{n} + \frac{n(n - 1)}{2}\cdot \frac{1}{n^2} + \cdots + \frac{n(n - 1)\ldots(n - (n -1))}{n!} \cdot \frac{1}{n^n} $$  $$ \le 1 + \frac{1}{2!} + \cdots + \frac{1}{n!} =S_n $$  How can I show that $S_n \le $ a subsequence of $\left({1 + \frac 1 n}\right)^n$ or any sequence that converges to $e$?
118828
$$\sum_{n=1}^\infty \frac{1}{n^2 2^n}$$  I am new in series thus I tried a pair of methods to compute but I couldn't
87270
This comes from the comments section of this question , credits .  The statement is  $$\sum_{n=1}^\infty\left(e-\sum_{k=0}^n\frac1{k!}\right)=1$$  This looks really interesting, so I was wondering if anyone has any ideas/suggestions or even better the whole proof?  Disclaimer: I added that link specifically because this might look like homework, leading to "What have you tried?". So the disclaimer is, I actually haven't tried anything yet, because yes, asking the community is just easier. It would be great if someone could answer this though since I, personally at least, think that this equation is very interesting and adds value to any math enthusiast stumbling upon it here someday. :) I will myself be trying it meanwhile too though(it really is interesting) , and add and accept my solution if I am successful and there are no answers yet.     I tried  for $\sum_{k=0}^n\frac1{k!}$ and it yields  $$\sum_{k=0}^n\frac1{k!} = \frac{e\Gamma (n+1,1)}{\Gamma(n+1)}$$  where $\Gamma(a,r)$ is the incomplete gamma function and $\Gamma(a)$ is the euler gamma function. . Might be of help, might not. Just thought it'd be worth adding.
76113
If I were given $\lim_{n\to \infty}(1 + \frac{1}{n})^n$, and asked to solve, I would do so as follows:  $$\lim_{n\to \infty}(1 + \frac{1}{n})^n$$ $$=(1 + \frac{1}{\infty})^\infty$$ $$=(1 + 0)^\infty$$ $$=1^\infty$$ $$=1$$  I'm aware that this limit is meant to equal to $e$, and so I ask:     why don't I get $e$ when I solve $\lim_{n\to \infty}(1 + \frac{1}{n})^n$?
83864
In fact the real question is: Does $\lim\limits_{n\to\infty}1^{n}=e$?.  I know that   $$ \lim\limits_{n\to\infty}\left(1+\dfrac{1}{n}\right)^n=e, $$  So, can we say that $1^\infty=e$?  And, by logic, This product  $$\underset{\text{infinity times}}{1\cdot1\cdot\ldots\cdot1},$$ gives $1$ not other value.
100131
I want to calculate the sum:$$\sum_{n=1}^\infty\frac{1}{(n-1)!(n+1)}=$$  $$\sum_{n=1}^\infty\frac{n}{(n+1)!}=$$ $$\sum_{n=1}^\infty\frac{n+1-1}{(n+1)!}=$$ $$\sum_{n=1}^\infty\frac{n+1}{(n+1)!}-\sum_{n=1}^\infty\frac{1}{(n+1)!}=$$ $$\sum_{n=1}^\infty\frac{1}{n!}-\sum_{n=1}^\infty\frac{1}{(n+1)!}.$$  I know that $$\sum_{n=0}^\infty\frac{1}{n!}=e$$  so $$\sum_{n=1}^\infty\frac{1}{n!}=\sum_{n=0}^\infty\frac{1}{n!}-1=e-1$$  But, what can I do for $$\sum_{n=1}^\infty\frac{1}{(n+1)!}$$ ?  Am I allowed to start a sum for $n=-1$ ? How can I bring to a something similar to $$\sum_{n=0}^\infty\frac{1}{n!}$$?
55184
Deteremine the sum of the series $$\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \ldots$$  So I first write down the $n^{th}$ term $a_n=\frac{\frac{n(n+1)}{2}}{n!}=\frac{n+1}{2(n-1)!}$.  So from there I can write the series as $$1+\frac{3}{2}+\frac{4}{2\times 2!}+\ldots +\frac{n+1}{2(n-1)!}+\ldots $$  I am quite sure I can do some sort of term by term integration or differentiation of some standard power series and crack this. Any leads?
9797
When I first studied power series in high school, the teacher gave the following general definition:  \begin{equation} f(x)=\sum_{n=0}^{\infty}a_n (x-x_0)^n \end{equation}  He then proceeded to explain that when both $x=0$ and $x_0 =0$, the first term of the power series is defined to simply be $a_0$. How does one justify $0^0 = 1$ in this case?