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Whatever could be the $a_i$'s, you could sort them by increasing values and rewrite $$S=\sum_{i=1}^n e^{\lambda a_i}=\sum_{i=1}^n e^{\lambda b_i}$$ in which $b_1\leq b_2 \leq b_3\leq \cdots\leq b_n$. So, $$S=\sum_{i=1}^n e^{\lambda b_i}=e^{\lambda b_n}\sum_{i=1}^ne^{\lambda (b_i-b_n)}=e^{\lambda b_n}\Big(1+\sum_{i=1}^{n-1}e^{\lambda (b_i-b_n)}\Big)$$ So ...


1

The answer should be the largest of the $a_i$. To get to it intuitively from what you have in your last equation, note that if $a>b$, $e^{\lambda a}$ grows faster than $e^{\lambda b}$, since $e^{\lambda(a-b)} \to \infty$. You can use this principle to say that the top is dominated by the term where $a_i$ is largest, as is the bottom, so the limit tends to ...


3

It's, unfortunately, not a particularly well-defined problem, as infinite power towers aren't always well defined. However, if we want to apply algebraic techniques anyhow, notice that we can write it as $$2x^{\left(2x^{2x^{2x\ldots}}\right)}=4$$ where the inner expression on the left is equal to four for a solution, giving $$2x^4=4$$ which is easier to ...


2

$$ 2x^{2x^{2x^{..}}}=4 \longrightarrow 2x^4=4 $$ $$ x^4=2 $$ I think you can do the rest.


5

I guess the answer ("it is hard") have been given. The numerical computation gives $\sum_{n=0}^\infty{\mathrm e}^{-n^3}=1.368\dots$ I want to add that a natural first order approximation given by the Euler–Maclaurin formula, is $$\sum_{n=0}^\infty{\mathrm e}^{-n^a}\approx\frac{1}{2}+\Gamma\left(1+\frac{1}{a}\right)\approx 1.5-\frac{\gamma}{a},$$ where ...


5

No, such expression is not known in the world of commonly used special functions. Its closest analog is the sum $\sum\nolimits_{n=0}^{\infty}e^{-n^2}$, which is already quite nontrivial: it is expressed in terms of elliptic theta functions. One way to convince yourself that the sum is "too exotic" is to consider a generalization ...


2

Since: $$\sum_{n\geq 1}\frac{z^{n-1}}{\Gamma(n)\Gamma(n+1)}=\frac{I_1(2\sqrt{z})}{\sqrt{z}}$$ provided that $a,b\in\mathbb{R}^+$ you series equals: $$ a \int_{0}^{b}\frac{I_1(2\sqrt{az})}{\sqrt{az}}e^{-z}\,dz = \int_{0}^{ab}\frac{I_1(2\sqrt{z})}{\sqrt{z}}e^{-z/a}=2\int_{0}^{\sqrt{ab}}I_1(2u)\,e^{-u^2/a}\,du.$$


0

Let $$S=\sum_{k=1}^\infty \frac{k^2}{2^k}.$$ Shifting the indexes by one, we have $$S=\frac12+\sum_{k=1}^\infty \frac{(k+1)^2}{2^{k+1}}=\frac12+\frac12\left(\sum_{k=1}^\infty\frac{k^2+2k+1}{2^k}\right)=\frac12\left(1+S+2\sum_{k=1}^\infty\frac k{2^k}+1\right).$$ Let us now look at the auxiliary sum $$T=\sum_{k=1}^\infty \frac{k}{2^k}.$$ Again by shifting, ...


10

For $|r|<1,$ $$\sum_{k=0}^\infty r^k=\frac1{1-r}$$ Differentiate wrt $r$ and multiply by $r$ $$\sum_{k=0}^\infty kr^k=\frac{r}{(1-r)^2}$$ Again differentiate wrt $r$ and multiply by $r$ $$\sum_{k=0}^\infty k^2r^k=\frac{r(r+1)}{(1-r)^3}$$ Put $r=1/2$: $$\sum_{k=0}^\infty \frac{k^2}{2^k}=\frac{(1/2)(1/2+1)}{(1-1/2)^3}=6$$


1

We have $$\sum_{n\geq1}q^{-n^{2}}\cos\left(2nz\right)=\frac{1}{2}\left(\theta_{3}\left(z,\frac{1}{q}\right)-1\right)$$ then if we integrate with respect to $z$ we get $$\sum_{n\geq1}q^{-n^{2}}\frac{\sin\left(2nz\right)}{2n}=\frac{1}{2}\int\theta_{3}\left(z,\frac{1}{q}\right)dz-\frac{z}{2}.$$ Now you can make an estimation, because a I think there isn't a ...



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