I was constructing a proof through inequalities, but I am having a bit of problem showing the following step:  $$\sum_{k=N+1}^\infty \frac {1}{k!} &lt; \frac {1}{N!}$$   Is there any quick way to show this?
124186
$$\sum_{k=n}^\infty{\frac{1}{k!}} \leq \frac{2}{n!}$$  Can someone show why this estimate holds true? I tried quite a bit but couldn't really find a way to approach this.   says it is true but I don't know what the gamma function is.  $$ \sum_{k=n}^\infty{\frac{1}{k!}} = \frac{1}{n!} + \sum_{k = n+1}^\infty \frac{1}{k!}$$ So then I need to show that$$ \sum_{k=n+1}^\infty{\frac{1}{k!}} \leq \frac{1}{(n+1)!} ~~\Big[\leq \frac{1}{n!}\Big]$$  Is it possible to do this by induction? I don't really know how to approach this now.
123943
$$n!=\sum_{k=0}^n \binom{n}{k}(n-k+1)^n(-1)^k$$  Could anyone give the proof of the above equation? Thanks in advance!
96526
I have to show the inequality of $$\left(\sum_{i=1}^n x_i\right)*\left(\sum_{i=1}^n \frac{1}{x_i}\right) \geq n^2.$$For $x_1, ... x_n \in \mathbb{R_{&gt;0}}$ and $ n \geq 1$. I wanted to show this inequality by induction. The basis is clear, but I am not sure how to do the inductive step. Can someone help me with that?
111670
I have got an inequality problem which is as follow:     Show that $e^n&gt;\frac{(n+1)^n}{n!}$   I can do it by induction but I have been told to prove it without induction.  My Work:  $$e^n=1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+........$$ $$e^n&gt;1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+........+\frac{n^n}{n!}$$ $$e^n&gt;\frac{n^n}{n!}+\frac{n^{n-1}}{(n-1)!}.......+\frac{n^2}{2!}+n+1$$  From here I can't go further.   I shall be thankful if you guys can provide me a complete solution/proof of this inequality. A hint will also work.  Thanks in advance.
51322
Could please give me a hint.  $$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{k}{k^{2}+1}$$  It looks like an integral, but I failed to figure out the function.
52085
How can I find the value of the sum $\sum_{k=1}^{\infty}\frac{k^n}{k!}$? for example for $n=6$, we have $$\sum_{k=1}^{\infty}\frac{k^6}{k!}=203e.$$
26124
Can someone look at my proof. I am supposed to prove by induction. The question is to prove the following:  $$\sum _{i=0}^{n}{i} =\frac { n\left( n+1 \right)  }{ 2 } .$$  If $n=1$  Then  $$1=\sum_{i=1}^1{i}=\frac{1(1+1)}{2}=1$$  Now assume $n=k$.  Thus  $$\sum_{i=1}^k{i}=\frac{k(k+1)}{2}=1$$  If now $n=k+1$  $$\sum_{i=1}^{k+1}{i}=\sum_{i=1}^k{i}+(k+1)= \frac{(k+1)(k+2)}{2}$$  Is there anything more that I have to do, or is this it?
113204
How could the series below be computed ? $$\sum_{k=0}^{\infty}\frac{1}{2^{k!}}$$ It's not a series from a book, but a series I thought of many times, and I didn't manage to figure out what I should do here. I'm just curious to know if there are some known ways for approaching such a series. Thanks!
4063
I've been trying to prove   $$\sum_{k=1}^{n}{(-1)^{k+1} {{n}\choose{k}}\frac{1}{k}=H_n}$$  I've tried perturbation and inversion but still nothing. I've even tried expanding the sum to try and find some pattern that could relate this to the harmonic series but this just seems surreal. I can't find any logical reasoning behind this identity. I don't understand . Is there any proof that doesn't require the integral transform?
97681
I'm trying to prove that  $$\sum_{k=0}^n {n \choose k} (-1)^k \frac{1}{k+1} = \frac{1}{n+1}$$  So far I've tried induction (which doesn't really work at all), using well known facts such as  $$\sum_{k=0}^n {n \choose k} (-1)^k = 0$$  and trying to apply identities like  $${n \choose k} = {n-1 \choose k-1} + {n-1 \choose k}$$  Would anyone be able to point me towards the right method? Should I be looking to apply an identity or is there a method I'm missing?
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$$\sum_{n=1}^\infty \frac 1{n!}$$   I think we may compare it with $\frac{1}{n^n}$, and say that $\frac{1}{n!}&gt;\frac{1}{n^n}$. But I am not sure of the convergence or divergence of $\frac{1}{n^n}$ itself, so this turns out to be useless.  What other way is there?
118447
I have to give low and high bounds for the following:  $$ \sum_{n=1}^\infty \frac{1}{2^n - 3^n } $$  How do I determine an upper bound? How can I show this sum exists?  edit: removed erroneous conclusion.
42565
I've been dealing with this problem for almost 2 hours now, with hardly any progress. I'm to prove the following inequality using induction   $$\sqrt[n+1]{n+1}&lt;\sqrt[n]{n}$$ where $n≥3$, $n∈\mathbb{N}$. How can this be done?
91788
Prove that for all $n \ge 1$:  $$\sum_{k=1}^n \frac{1}{k(k+1)} = \frac{n}{n+1}$$  What I have done currently:  Proved that theorem holds for the base case where n=1.  Then:  Assume that $P(n)$ is true. Now to prove that $P(n+1)$ is true:  $$\sum_{k=1}^{n+1} \frac{1}{k(k+1)} = \frac{n}{n+1} + n+1$$  So: $$\sum_{k=1}^{n+1} \frac{1}{k(k+1)} = \frac{n+1}{n+2}$$  However, how do I proceed from here?
113627
$a_n=(1+\frac{1}{n})^n$ ,  $b_n=\sum_{k=0}^n \frac{1}{k!}$.  Show that $b_n-\frac{3}{2n} &lt; a_n &lt; b_n$.
121563
Let $p&gt;1$. Prove that the series:  $\sum_{n=1}^\infty \frac {(-1)^{n+1}}{n^p}$  is between $\frac {1}{2}$ and 1.  Any help is appreciated.  Just a challenge problem I was presented and curious on the solution  Thanks
114099
Given the Series $$\sum_{k=1}^\infty \frac{1}{k(k+2)}$$  How exactly would I find out the limit is $\frac34$ as suggested by Wolframalpha? I already found out I can prove it actually converges by performing the comparison test and seeing that the underlying sequence isn't a null-sequence. But unfortunately I am absolutely clueless on how to prove that it converges to $\frac34$.  Regards,  Dennis