I know that $e^xe^y=e^{x+y}$ but I want to show it by expanding the exponentials in MacLaurin Series.  $$ \left(\sum_{n=0}^{\infty} \frac{x^n}{n!}\right) \left(\sum_{m=0}^{\infty} \frac{y^m}{m!}\right) =^? \sum_{n=0}^{\infty} \frac{(x+y)^n}{n!} $$  This is what I have.  I'm not sure what my next step would be since I can't think of a way to combine the two summations on the RHS.
31145
In order to show $e^{x+y}=e^{x}e^{y}$ by using Exponential Series, I got the following:   $$e^{x}e^{y}=\Big(\sum_{n=0}^{\infty}{x^n \over n!}\Big)\cdot \Big(\sum_{n=0}^{\infty}{y^n \over n!}\Big)=\sum_{n=0}^{\infty}\sum_{k=0}^n{x^ky^n \over {k!n!}}$$  But, where should I go next to get $e^{x+y}=\sum_{n=0}^{\infty}{(x+y)^n \over n!}$.  Thanks in advance.
80433
By expanding $e^x$ into a series $\sum e^x$ prove that $$\forall x \in \mathbb{R}, x \ge 0 \implies e^{x-1} \ge x$$  Also show when this inequality becomes equality.  I'm not really sure how to attack this problem at all, any tips would be very welcome.
120508
Given $a_0 = 1$ and $a_n = (n + 1)a_{n - 1} + 3^n$ for $n \geq 1$, find formula for $a_n$.   First remind the definition of power series $F(x) = \sum_{n \geq 0}f_n \frac{x^n}{n!}$ and exponential generating function $G(x) = \sum_{n \geq 0}\frac{x^n}{n!} = e^x$   Note we have to use exponential generating function to solve this question. So let $f(x) = \sum_{n \geq 0}a_n\frac{x^n}{n!}$, multiplying both sides of  $a_n = (n + 1)a_{n - 1} + 3^n$ by $\frac{x^n}{n!}$ and sum over all $n \geq 1$. We get, $$\sum_{n \geq 1}a_n\frac{x^n}{n!} = \sum_{n \geq 1}(n + 1)a_{n - 1}\frac{x^n}{n!} + \sum_{n \geq 1}3^n\frac{x^n}{n!}$$ $$\sum_{n \geq 1}a_n\frac{x^n}{n!} = \sum_{n \geq 1}na_{n - 1}\frac{x^n}{n!} + \sum_{n \geq 1}a_{n - 1}\frac{x^n}{n!} + \sum_{n \geq 0}\frac{(3x)^{n}}{n!} - 1$$ $$\underbrace{\sum_{n \geq 0}a_n\frac{x^n}{n!}}_{f(x)} - 1 = x\underbrace{\sum_{n \geq 1}a_{n - 1}\frac{x^{n-1}}{(n-1)!}}_{f(x)} + \underbrace{\sum_{n \geq 1}a_{n - 1}\frac{x^{n}}{(n)!}}_{\int{f(x)} = F(x) + C} + e^{3x} - 1$$ $$F'(x) - 1 = xF'(x) + F(x) + e^{3x} - 1   + C$$ $$F'(x)(1 - x) - F(x) = e^{3x} + C$$ I think I'm essentially stuck here, I'm trying my best to manipulate each term so that we can express each term in terms of $f(x)$. But I can't go further from here because it doesn't like look this form can lead us to the solution.   Any feedback or suggestion?
25721
$$\sum_2^\infty e^{3-2n}$$  The formulas for these things are so ambiguous I really have no clue on how to use them.  $$\frac {cr^M}{1-r}$$  $$\frac {1e^2}{1-\frac{1}{e}}$$  Is that a wrong application of the formula and why?
73042
Here's the problem: Let $x$ be any real number. Show that $$ \lim_{m \to \infty} \left( 1 + \frac{x}{m} \right)^m = \sum_{n=0}^ \infty \frac{x^n}{n!}  $$ I'm sure there are many ways of pulling this off, but there are 3 very important hints to complete the exercise in the desired manner:   Expand the left side as a finite sum using the Binomial Theorem. Call the summation variable $n$. Now add into the finite sum extra terms which are $0$ for $n&gt;m$, in order to make it look like an infinite series. What happens to the limit on $m$ outside the series?   So far I was able to use Hint 1 to expand the left side: $$ \lim_{m \to \infty} \left( 1 + \frac{x}{m} \right)^m = \lim_{m \to \infty} \sum_{n=0}^m \binom {m}{n} \left( \frac{x}{m} \right)^n $$  No matter what I do with the binomial coefficients and factorials, I can't figure out what extra terms to add per Hint 2. Any suggestions?
30300
Maclaurin series for   $$\frac{x}{e^x-1}$$  The answer is   $$1-\frac x2 + \frac {x^2}{12} - \frac {x^4}{720} + \cdots$$  How can i get that answer?
21104
Hello I have some problems concerning Taylor series.  Given the function $$f(x)=e^{\sin{x}} $$  I concluded that the Taylor series expansion would be   $$f(x) = \sum^\infty_{n=0}\frac{1}{n!}f^{(n)}(x)(x-x_0)^n$$ But Wolfram wrote a much simpler form   $$f(x)=\sum^\infty_{n=0}\frac{\sin^k{x}}{k!}$$  My question is: how come? Secondly, could somebody explain to me how  to get the second sum's radius of convergence?   Thank you very much for your time!
5291
Compute : $$L=\lim_{x\to 1^-} \left(\prod_{n=0}^{\infty} \left(\frac {1+x^{n+1}}{1+x^n}\right)^{x^n}\right)$$   My try:   Writing out the first few terms I noticed that the limit can be expressed as $$L=\lim_{x\to 1^-} \frac 12\left(\prod_{n=1}^{\infty} (1+x^n)^{x^{n-1} -x^n}\right)$$ $$L=\frac 12\lim_{x\to 1^-} \left(\prod_{n=1}^{\infty} (1+x^n)^{x^{n-1}}\right)^{1-x}$$  Converting to exponential form I get $$L=\frac 12 \exp {\left(\lim_{x\to 1^-} (1-x)\sum_{n=1}^{\infty} \left(x^{n-1}\ln(1+x^n)\right)\right)}$$  And got stuck here. I did try to use approximation $\ln x\sim x$ hereon to get final answer as $\frac {\sqrt e}{2}$ but I think it was improper to use approximation and hence I believe that the answer I got is flawed. I also noticed that the form I obtained is quite similar to be used for a Riemann Sum but I didn't get a way to tackle the problem using that way.  Can someone please help me with this problem....
116995
$$\lim_{x\rightarrow +\infty}\sqrt{x}e^{-x}\left(\sum_{k=1}^{\infty}\frac{x^{k}}{k!\sqrt{k}}\right)$$ Any hint will be appreciated.  Note: There is a related question on MathOverflow: . The MO question references this question and it also links to some other posts containing similar expressions.
114880
We have the sum:  $\sum\nolimits_{n=0}^\infty \frac{n}{n!}$  Why is this sum a representation of $e$ when the sum:  $\sum\nolimits_{n=0}^\infty \frac{1}{n!}$  is already a representation of $e$?  For me it seams that the first row is greater then the second.
62560
$$ \sum_{n=1}^\infty \frac{n!e^n}{n^n} $$ Where $e$ is Euler's number.  Recently on calculus class we were covering convergence tests and my group got stuck with this infinite series. Our calculus teacher told us that it diverges but he doesn't see any elementary way or simple test to prove it. He told us something about Euler's Gamma function and said that he knows it diverges because of Stirling's formula but this is not quite our level yet.  I tried to think about it myself but I established only that this  is some kind of an edge case. We could generalize the problem to series of $\frac{n!a^n}{n^n}$. Then, from the ratio test we know that it converges for $a &lt;e$ and diverges for $a&gt;e$. For $a=e$ the test is inconclusive. I've also tried to compare it with various divergent series, but nothing worked...  Does anybody have any idea for a simple solution?
101345
I tried getting it into a closed form but failed. Could someone help me out?  $$\sum_{n=0}^{\infty} \dfrac{a^n}{(3n)!}$$
36658
I know that $$ \int_0^\infty \frac{\log x}{\exp x} = -\gamma $$ where $ \gamma $ is the Euler-Mascheroni constant, but I have no idea how to prove this.  The series definition of $ \gamma $ leads me to believe that I should break the integrand into a series and interchange the summation and integration, but I can't think of a good series. The Maclaurin series of $ \ln x $ isn't applicable as the domain of $ x $ is not correct and I can't seem to manipulate the integrand so that such a Maclaurin series will work.   Another thing I thought of was using $ x \mapsto \log u $ to get $ \int\limits_{-\infty}^\infty \frac{\log \log u}{u^2} \ du $ and use some sort of contour integration, but I can't see how that would work out either.
63519
$$\lim_{n\to\infty} (1+\frac{2}{n})^{n^2} e^{-2n} = ? $$         $e$   $e^2$   $e^{-1}$   $e^{-2}$      My answer is 1. Since $$\lim_{n \to \infty} \left(1+\frac{2}{n}\right)^{n} =e^{+2}.$$  Therefore $$\lim_{n\to\infty} \left(1+\frac{2}{n}\right)^{n^2} =e^{2n}.$$  Where am I wrong?
33255
Let's say I want to use Maclaurin series to get the series expansion $S(x)$ for $f(x) = e^x$ where $S(x) = c_0x^0 + c_1x^1 + c_2x^2 + c_3x^3 + ...$  $f(0) = S(0) = c_0 = e^0 = 1$ so that's fine.  $f'(0) = S'(0) = c_1 = e^0 = 1$ again, fine.  $f''(0) = S''(0) = 2c_2 = e^0 = 1$ so $c_2 = 1/2$, fine.  $f'''(0) = S'''(0) = 6c_3 = e^0 = 1$ so $c_3 = 1/6$, fine.  $f''''(0) = S''''(0) = 24c_4 = e^0 = 1$ so $c_4 = 1/24$, fine.  And so on, so we conclude that $c_k = \frac{1}{k!}$ so then:  $$S(x) = \sum_{k=0}^{\infty} \frac{x^k}{k!}$$  But now what I don't understand is what allows us to go "Furthermore, $S(x) = f(x)$ for all $x$"! In fact what makes it valid for us to plug any other number other than $x=0$ into this? I see this equation used as a straight-up equivalent to $e^x$ even though we used $x=0$ and nothing else.  So I thought I would try it with a neighborhood of $1$ instead to see what happens:  $S(x) = c_0(x-1)^0 + c_1(x-1)^1 + c_2(x-1)^2 + c_3(x-1)^3 + ...$  $f(1) = S(1) = c_0 = e^1 = e$ so that's fine.  $f'(1) = S'(1) = c_1 = e^1 = e$ again, fine.  $f''(1) = S''(1) = 2c_2 = e^1 = e$ so $c_2 = e/2$, fine.  $f'''(1) = S'''(1) = 6c_3 = e^1 = e$ so $c_3 = e/6$, fine.  $f''''(1) = S''''(1) = 24c_4 = e^1 = e$ so $c_4 = e/24$, fine.  Looks pretty similar:  $$S(x) = \sum_{k=0}^{\infty} \frac{e(x-1)^k}{k!}$$  and, in fact, for a neighborhood of $x=a$:  $$S(x) = \sum_{k=0}^{\infty} \frac{e^a(x-a)^k}{k!}$$  Now we have $e$ inside the function expansion for $e^x$ which seems a little circular?  I guess my question is this: We created the function $S$ to have the same-valued $n$th derivatives as $f(x)$ but only at $x=0$ or $x=1$, or $x=a$, etc, but the series representations look different depending on which neighborhood I pick, and in some cases the expansions seem to include the very number we're trying to describe.  How do we know which is "right" or that it is even "right" to use for all $x$? I know the usual response to this is that it's equivalent when $f(x)$ is analytic but that doesn't help me at all because it says the function is analytic when the Taylor series around $x_0$ converges to the function in a neighborhood around $x_0$ for all $x_0$ in the function's domain. But how do I know this?  Yet again feels circular... just working through these two examples makes me wonder, how do I know these series converge to the function itself? How do I know they're equivalent representations? Am I supposed to avoid the self-referencing?  I'm looking for some context behind how to make sense of these two Taylor series and how I am supposed to know that $e$ is analytic or that I can use $S(x)$ for any $x$ I want even if I only computed it for the neighborhood around $x=0$.
20827
I need to find a function $f(x)$ which is equal to the sum $$ \sum_{n=0}^\infty \frac{n}{n+1}x^n, $$ for every $x\in \mathbb{R}$ for which the series converge.  Now, using WolframAlpha, I've found the answer  $$f(x)=\frac{-x+x\ \text{log}(1-x)-\text{log}(1-x)}{(x-1)x},\ \ \ \text{when}\ |x|&lt;1$$ However I'm trying to figure out how to find this function $f(x)$ on my own. I've tried to combine the Maclaurin series of $\ln(1+x)$, $\sin(x),$ and several other functions but I don't seem to succeed. Any suggestions how to work out this problem from scratch?
90299
My question is too simple. We know all that if we define the exponential function on $\mathbb{C}$ then we define the real part and imaginary part of $\exp{it}$ as $\cos{t}$ and $\sin{t}$.  So if we assume that we know the definition of the trigonometric functions and the exponential function not as a series but for exponential as solution of differential equation $y'-y=0$ then my question is     How can I explain this relation (intrinsically)in a simple way to a high-school student, they is any physical interpretation of this formula  ?   or we  can give another equivalent question (in my sens)     How can one prove the Euler relation? and What was the original proof of Euler himself?