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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 added 1121 characters in body 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 added 76 characters in body 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 added 17 characters in body 1d revised Can I get better approximation of $\sum_{k=1}^{n} k^k$ added 1 character in body 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 awarded Custodian 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?