Find $f(x)$, the unknown function satisfying $$f(x) = \sum\limits_{n=0}^{+\infty}\frac{x^{4n}}{(4n)!}$$ I’m looking for a direct solution which is different from mine, if possible.

Let $(x_n)$ a decreasing sequence and $\sum x_n\to s$. Then $(n\cdot x_n)\to 0$ Check my proof please, Im not completely sure about it correctness. If $\sum_{k=h}^\infty x_k= s$ then we can rewrite the sum for starting index $1$ with the change $k-h=j$, then $$\left(\sum_{j=1}^n x_j\right)-n\cdot x_n=\sum_{j=1}^n (x_j-x_n)$$ Then taking limits $$\color{red}{\lim_{n\to\infty}\left[\left(\sum_{j=1}^n x_j\right)-n\cdot x_n\right]}=\lim_{n\to\infty}\sum_{j=1}^n (x_j-x_n)=\sum_{j=1}^\infty (x_j-0)=\color{red}{\lim_{n\to\infty}\sum_{j=1}^n […]

Convergence in probability implies convergence on a subsequence almost surely. But this means we fix a subsequence, such that $X_{n_k}$ converges for almost every $\omega$, right? The subsequence we pick does not depend on the $\omega$ right?

This question already has an answer here: Vandermonde's Identity: $\sum_{k=0}^{n}\binom{R}{k}\binom{M}{n-k}=\binom{R+M}{n}$ 4 answers

How may one calculate $$\lim_{n\to\infty} \ \left(\left(\sum_{k=1}^{n} \frac{1}{3k-1}\right) – \frac{\ln n}{3}\right) \ ?$$

Find the value of $$S=\sum_{n=1}^{\infty}\left(\frac{2}{n}-\frac{4}{2n+1}\right)$$ My Try:we have $$S=2\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{2}{2n+1}\right)$$ $$S=2\left(1-\frac{2}{3}+\frac{1}{2}-\frac{2}{5}+\frac{1}{3}-\frac{2}{7}+\cdots\right)$$ so $$S=2\left(1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\cdots\right)$$ But we know $$\ln2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$$ So $$S=2(2-\ln 2)$$ Is this correct?

Physicists often assign a finite value to a divergent series $\sum_{n=0}^\infty a_n$ via the following regularization scheme: they find a sequence of analytic functions $f_n(z)$ such that $f_n(0) = a_n$ and $g(z) := \sum_{n=0}^\infty f_n(z)$ converges for $z$ in some open set $U$ (which does not contain 0, or else $\sum_{n=0}^\infty a_n$ would converge), then […]

How to find a closed form for the expression?? $$ 1+\frac 1 2 +\frac{1 \times2}{2 \times 5}+\frac{1 \times2\times 3}{2 \times5\times 8}+\frac{1\times 2\times 3\times 4}{2 \times 5\times 8\times 11}+ \cdots$$ Wolfram alpha gives, $$\frac{3}{2}+\frac{\ln(\sqrt[3]{2}-1)}{4\sqrt[3] {2}}+\frac{\sqrt{3}}{2\sqrt[3]{2}}\arctan\frac{\sqrt{3}}{2\sqrt[3]{2}-1}$$

Questions that ask for “intuitive” reasons are admittedly subjective, but I suspect some people will find this interesting. Some time ago, I was struck by the coincidence that the Euler-Mascheroni constant $\gamma$ is close to the square root of 1/3. (Their numerical values are about 0.57722 and 0.57735 respectively.) Is there any informal or intuitive […]

How to prove that the sequence $a_n=n^{1/n}$ is convergent using definition of convergence?

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