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1,809
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Question on transforming a sum to an integral using the Euler–Maclaurin formula.
I have a question regarding transforming a summation to an integral using the Euler–Maclaurin formula. Imagine I have this sum
$$\sum_{i=0}^{m} f(i) \qquad \text{with} \qquad f(i)= \exp [(\frac{a}{b+...
- 51
1
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1
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123
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Show that $\left | \sum_{k=1}^\infty \frac{\left(e^{ik} + \sin(\sqrt 2 k) + \cos(\sqrt 3 k)\right)^3}{k} \right| < 6$
I am trying to show the following sum is bounded:
$$ \sum_{k=1}^\infty \frac{\left(e^{ik} + \sin(\sqrt 2 k) + \cos(\sqrt 3 k)\right)^3}{k}$$
and to show that the magnitude
$$\left | \sum_{k=1}^\infty \...
- 8,369
1
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0
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69
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Leibniz integral rule for summation
Context
Fundamental points of Feymann trick:
You have an integral $I_0=\int_a^b f(t)\mathrm{d}t$
Now consider a general integral $I(\alpha)=\int_a^b g(\alpha,t)\mathrm{d}t$ so that $I'(\alpha)=I_0$ ...
10
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3
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190
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Show that $\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^2} \sum_{n=1}^k \frac{1}{n}=\frac{5\zeta(3)}{8}$
$$\sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{k^2} \sum_{n=1}^k \dfrac{1}{n}=\frac{5\zeta(3)}{8}$$
I tried to create a proof from some lemmas some are suggested by my Senior friends
Lemma 1 $$
{H_n} = \sum\...
- 45.4k
2
votes
2
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99
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Prove $\frac12\left(\psi\left(\frac{x+1}2\right)-\psi\left(\frac x2\right)\right)=\psi(x)-\psi\left(\frac x2\right)-\ln2$
Desmos suggests that$$\frac12\left(\psi\left(\frac{x+1}2\right)-\psi\left(\frac x2\right)\right)=\psi(x)-\psi\left(\frac x2\right)-\ln2$$Where $\psi$ is the digamma function. I can write the LHS as $$\...
- 6,549
2
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0
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67
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Closed form for $\psi^{1/k}(1)$, where $k$ is an integer
I have proven the identity
$$
\sum_{k=1}^{\infty} \dfrac{\operatorname{_2F_1}(1, 2, 2-1/t,-1/k)}{{k}^{2}} = Γ(2-\dfrac{1}t){\psi^{1/t}(1)}+\psi(-\dfrac{1}t)(\dfrac{1}t(1-\dfrac{1}t))+\gamma(1-\dfrac{1}...
0
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0
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104
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Is this divergent series, convergent?
Examining the series $\sum_{n=1}^{\infty} \frac{1}{nx}$ alongside its integral counterpart reveals insights into its convergence. Notably, the integral over intervals from $10^n$ to $10^{n+1}$ yields ...
- 11.2k
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2
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82
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Upper rectangle area sum to approximate 1/x between $1\leq x\leq 3$
I am trying to figure out how to use rectangles to approximate the area under the curve $1/x$ on the interval $[1,3]$ using $n$ rectangle that covers the region under the curve as such.
Here is what I ...
- 2,747
0
votes
3
answers
2k
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Finding the formula for the summation $\sum_{i=1}^n (2i-1) = 1+3+5+...+(2n-1)$.
I'm going through Calculus by Spivak and one of the questions is to find a formula for a summation.
$$\sum_{i=1}^n (2i-1) = 1+3+5+...+(2n-1).$$
I got the correct answer, $n^2$, but did it an ...
CommunityBot
- 1
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2
answers
80
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On the convergence of $\sum_{n=0}^\infty(f(x)-T_n\{f\}(x))$
To test the efficiency of the Taylor Series at approximating functions, I was wondering whether$$\sum_{n=0}^\infty(f(x)-T_n\{f\}(x))\tag{$\star$}$$converges, where $T_n\{f\}(x)$ is the degree $n$ ...
- 6,549
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0
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31
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Can we compare the arithmetic mean of the ratios given the comparison between individual arithmetic means?
I have positive real random numbers $u_1,\ldots,u_n$ and $v_1,\ldots,v_n$ and $x_1,\ldots,x_n$ and $y_1,\ldots,y_n$.
I know that the arithmetic mean of $u_i$'s is greater than the arithmetic mean of $...
- 452
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0
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135
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Simplify a summation in the solution of $\displaystyle\int_{0}^{\infty}e^{-cx}x^{n}\arctan(ax)\mathrm{d}x$
Context
I calculated this integral:
$$\begin{array}{l}
\displaystyle\int_{0}^{\infty}e^{-cx}x^{n}\arctan(ax)\mathrm{d}x=\\
\displaystyle\frac{n!}{c^{n+1}}\left\lbrace\sum_{k=0}^{n}\left[\text{Ci}\left(...
1
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1
answer
95
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Evaluation of $\int_{0}^{\frac{\pi}{4}} \frac{\log(\log(\tan(\frac{\pi}{4} + x))) \cdot \log(\tan(\frac{\pi}{4} + x)))}{\tan(2x)} \,dx$ [closed]
$$\int_{0}^{\frac{\pi}{4}} \frac{\log(\log(\tan(\frac{\pi}{4} + x))) \cdot \log(\tan(\frac{\pi}{4} + x)))}{\tan(2x)} \,dx$$
$$\int_{0}^{\frac{\pi}{4}} \frac{\log(\log(\tan(\frac{\pi}{4} + x))) \cdot \...
5
votes
2
answers
157
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Show that $\sum_{n=1}^{\infty} \frac{\binom{2n}{n} (H_{2n} - H_n)}{4^n (2n - 1)^2} = 2 + \frac{3\pi}{2} \log(2) - 2G - \pi$
Show that $$\sum_{n=1}^{\infty} \frac{\binom{2n}{n} (H_{2n} - H_n)}{4^n (2n - 1)^2} = 2 + \frac{3\pi}{2} \log(2) - 2G - \pi$$
My try :
We know that
$$\sum_{n=1}^{\infty} \binom{2n}{n} (H_{2n} - H_{n}) ...
- 34.2k
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365
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A sum of two curious alternating binoharmonic series
Happy New Year 2024 Romania!
Here is a question proposed by Cornel Ioan Valean,
$$\sum_{n=1}^{\infty}(-1)^{n-1} \frac{1}{2^{2n}}\binom{2n}{n}\sum_{k=1}^n (-1)^{k-1}\frac{H_k}{k}-\sum_{n=1}^{\infty}(-1)...
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