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13h
comment Convergence of $\sum_{n=1}^{\infty} \log~ ( n ~\sin \frac {1 }{ n })$
Yes.. using the Taylor expansion of ${\sin x \over x}$.
13h
answered Convergence of $\sum_{n=1}^{\infty} \log~ ( n ~\sin \frac {1 }{ n })$
Jan
20
revised Prove or disprove: $\sum a_n$ convergent, where $a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}$.
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Jan
19
answered Prove or disprove: $\sum a_n$ convergent, where $a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}$.
Jan
19
comment How can I prove that $x-{x^2\over2}<\ln(1+x)$
You have to plug in $x = 0$ too
Jan
19
answered If $f,k\in C^{2\pi}$, then $\int_{-\pi}^{\pi}f(x+t)k(t)\mathop{dt}\in C^{2\pi}$
Jan
19
awarded  algebra-precalculus
Jan
18
comment Showing that $\|\sigma_n(f)\|_2\leq \|f\|_2$
The $k$th Fourier coefficient of $\sigma_n(f)$ is $(1 - {k \over n + 1})$ times the $n$th Fourier coefficient of $f$, so Bessel's inequality is enough too. (Keep in mind the explicit formulas for the $L^2$ norms of trig polynomials in terms of their Fourier coefficients.)
Jan
18
comment Write the proposition in words - $\urcorner\left(\forall x P\left(x\right)\right)$
Hint: $\urcorner \forall$ is the same as $\exists \urcorner$.
Jan
18
comment Showing that $\|\sigma_n(f)\|_2\leq \|f\|_2$
Plancherel's theorem gives this immediately since the magnitude of any Fourier coefficient of $\sigma_n(f)$ is always at most that of the corresponding Fourier coefficient of $f$
Jan
18
comment Calculate $ S =\sum_{k=1}^n\frac {1}{k(k+1)(k+2)}. $
This should be right now.
Jan
18
revised Calculate $ S =\sum_{k=1}^n\frac {1}{k(k+1)(k+2)}. $
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Jan
18
comment Calculate $ S =\sum_{k=1}^n\frac {1}{k(k+1)(k+2)}. $
Corrected, thanks
Jan
18
revised Calculate $ S =\sum_{k=1}^n\frac {1}{k(k+1)(k+2)}. $
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Jan
18
answered Calculate $ S =\sum_{k=1}^n\frac {1}{k(k+1)(k+2)}. $
Jan
18
comment Is the converse of Sylvester's inertia law true?
If two matrices are conjugate to a third, they are conjugate to each other.
Jan
13
answered Proving $proj_{proj_{\vec u} \vec v} \vec v=proj_{\vec u} \vec v$
Jan
10
comment Limit of S_n, where $S_n = \int_{0}^{1}\frac{nx^{n-1}}{1+x} dx$ for $n\ge 1$
I just noticed $nx^{n-1}$ integrates to $x^n$ so integrating by parts seemed like a good idea... it gets rid of the $n$ factor at least. I don't really know if there's a source that systematically treats such integrals
Jan
10
comment Limit of S_n, where $S_n = \int_{0}^{1}\frac{nx^{n-1}}{1+x} dx$ for $n\ge 1$
You've just shown the limit, if it exists, is at most 1.
Jan
10
answered Limit of S_n, where $S_n = \int_{0}^{1}\frac{nx^{n-1}}{1+x} dx$ for $n\ge 1$