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1d
comment Asymptotic of an interesting recurrence relation
@Jack Sorry for being too concise. I used to think that $\log{n}+1$ is almost the same, but actually it needs more estimates.
1d
revised Asymptotic of an interesting recurrence relation
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1d
comment Some properties about the Kampé de Fériet function
@HarryPeter Just consider those with same $s_1+\ldots+ s_n$. And notice the multinomial theorem! See here
1d
comment Prove $\{s_n\}$ converges if $\{a_n = s_n + 2s_{n+1}\}$ converges.
Okay, it's simple when you just compute the first three terms. $a_0=s_0+2s_1$ thus $s_1=\frac{1}{2}(a_0-s_0)$ and $s_2=\frac{1}{2}(a_1-s_1)=\frac{1}{2}a_1-\frac{1}{4}a_0+\frac{1}{4}s_0$... Finding the pattern will give you the answer, at least in this simple case
1d
accepted Can I get better approximation of $\sum_{k=1}^{n} k^k$
1d
answered Prove $\{s_n\}$ converges if $\{a_n = s_n + 2s_{n+1}\}$ converges.
1d
comment Prove $\{s_n\}$ converges if $\{a_n = s_n + 2s_{n+1}\}$ converges.
@Umakant Actually, we may have$$s_n=\sum_{k=0}^{n-1} (-2)^k a_{n-k}+(-2)^n s_0.$$ For instance, we let $a_n=2^{-n}$. Then $$s_n=2^{-n}\frac{(-4)^n-1}{-5}+(-2)^{s_0}.$$ When $s_0\not=\frac{1}{5}$, the sequence will diverge.
1d
comment How to get this inequality
@20824 Sorry for my mistake. Corrected!
1d
revised How to get this inequality
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1d
comment Prove $\{s_n\}$ converges if $\{a_n = s_n + 2s_{n+1}\}$ converges.
$s_n=(-2)^n$,then $a_n=(-2)^n+2(-2)^{n-1}=0$,but $s_n$ is not convergent..
1d
revised Problem on Euler's Phi function
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1d
revised Can I get better approximation of $\sum_{k=1}^{n} k^k$
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1d
answered Asymptotic of an interesting recurrence relation
1d
answered Trig and Geometry problem
2d
comment Prove that $\int_0^1\int_x^1 \frac{f(y)}ydy\,dx=\int_0^1f(x)\,dx$ if $f$ is Lebesgue integrable
$\int_0^1 f(x)dx$ is the same as $\int_0^1 f(y)dy$. It doesn't matter which symbol you use. For instance $\int_0^1 x^2+e^xdx=\int_0^1 y^2+e^ydy$
2d
comment Prove that $\int_0^1\int_x^1 \frac{f(y)}ydy\,dx=\int_0^1f(x)\,dx$ if $f$ is Lebesgue integrable
But I don't see the difference.. What's your question?
2d
comment Prove that $\int_0^1\int_x^1 \frac{f(y)}ydy\,dx=\int_0^1f(x)\,dx$ if $f$ is Lebesgue integrable
It's just Fubini's theorem. By tracking the region of integration, you will find that the region is exactly the triangle with vertices $(0,0),(1,1)$ and $(0,1)$. Thus integrating first via $x$, we will have $x$ integrating from $0$ to $y$.
2d
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2d
answered Prove that $\int_0^1\int_x^1 \frac{f(y)}ydy\,dx=\int_0^1f(x)\,dx$ if $f$ is Lebesgue integrable
2d
reviewed Approve Which discrete mathematics book do you think is better between Epp's and Rosen's for a clueless self-learner?