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May
8
comment Curvature for tautological bundle of projectivation
Ok! I've got your point, thanks!
May
8
accepted Curvature for tautological bundle of projectivation
May
8
comment Curvature for tautological bundle of projectivation
Thanks! So if I use $h(z,z)=\sum_{i,j} h^{ij} z_i \bar{z}_j$, then the holomorphic section is then $\frac{1}{h^{jj} z_j \bar{z_j}}\sum_{i,j} h^{ij} z_i \bar{z}_j$?
May
8
revised Curvature for tautological bundle of projectivation
added 52 characters in body
May
8
asked Curvature for tautological bundle of projectivation
May
8
reviewed Reject Find the limit of a sequence defined by $x_n =\sqrt{n^2+1}-n, n\in\mathbb{N}$
Apr
27
reviewed Approve Discrete Mathematics Proof Question
Apr
27
reviewed Reject An intuitive definition of contour integration.
Apr
27
reviewed Approve general formula of a sequence
Apr
27
answered Linear Operator Boundedness
Apr
25
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.
Apr
25
revised Asymptotic of an interesting recurrence relation
added 1121 characters in body
Apr
25
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
Apr
25
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
Apr
25
accepted Can I get better approximation of $\sum_{k=1}^{n} k^k$
Apr
25
answered Prove $\{s_n\}$ converges if $\{a_n = s_n + 2s_{n+1}\}$ converges.
Apr
25
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.
Apr
25
comment How to get this inequality
@20824 Sorry for my mistake. Corrected!
Apr
25
revised How to get this inequality
added 76 characters in body
Apr
25
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..