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## Hot answers tagged power-series

2

You're right. This series defines function: $$e^{3x}=\sum_{n=0}^{\infty}\frac{3^nx^n}{n!}$$ So it's convergent everywhere.There's something wrong with answer $\frac{1}{3}$.

2

Hint: $\sum\frac{(it)^k}{k!} a^{2k+1}=a\sum \frac{(ita^2)^k}{k!}$. What is $\sum \frac{x^k}{k!}$ (note summation starts from $k=1$, not $k=0$)? Then plug $x=ita^2$.

2

As Semiclassical noted, you look at $$\sum_{n=0}^\infty \dfrac{r^2 - (2n+1)^2}{(r^2 + (2n+1)^2)^2}$$ Expand the summand in partial fractions: $$\dfrac{r^2 - (2n+1)^2}{(r^2 + (2n+1)^2)^2} ={\frac {2 {r}^{2}}{ \left( r^2 + (2n+1)^2 \right) ^{2}}} - \dfrac{1}{ r^2 + (2n+1)^2 }$$ First deal with the term on the right: $$F(r) = \sum_{n=0}^\infty ... 1 By the ratio test we get$$\frac{a_{n+1}}{a_n}=3\frac{\left(\frac23\right)^{n+1}+1}{\left(\frac23\right)^{n}+1}\xrightarrow{n\to\infty}3\implies R=\frac13$$Remark Notice that if \sum a_n x^n and \sum b_n x^n are two power series with radius of convergence R_a and R_b and if R_a\ne R_b then the radius of convergence of \sum (a_n+b_n)x^n is ... 1 A start: We have for suitable x$$\frac{1}{1-x}=1+x+x^2+x^3+x^4+\cdots.$$Differentiate twice. 1 The root test is the correct idea here. To see which root you need to take, though, keep in mind that you could 'refill' your sum with the missing powers of x by filling in a coefficient of 0, thus we have \sum_{k=1}^{\infty}{\frac{x^{2k-1}}{2k-1}}=\sum_{k=1}^{\infty}{x^na_n}, where a_k=\frac{1}{k} for odd k and a_k=0 for even k. Now the root ... 1 The function u(x) to be expanded is such that u(0)=0 and$$u'(x)=\frac1{2x}\left(\frac1{\sqrt{1-x}}-1\right).$$Assuming one knows that$$\frac1{\sqrt{1-x}}=\sum_{n\geqslant0}a_nx^n,$$with a_0=1, this yields$$u(x)=\sum_{n\geqslant0}\frac{a_n}{2n}x^n.$$Can you identify (a_n)_{n\geqslant1}? Hint:$$\frac1{\sqrt{1-x}}=(1-x)^\alpha,\qquad ...

1

If $q>1$ then let $q'$ such that $q>q'>1$ and since $$(\ln k)^p=_\infty o(k^{q'})$$ then we see that the given series is absolutely convergent by comparison with a Riemann series. If $0<q\le1$ and using the function $$f(x)=\frac{(\ln x)^p}{x^q}$$ so $$f'(x)=\frac{p x^{q-1}(\ln x)^{p-1}-qx^{q-1}(\ln x)^p}{(x^q)^2}=\frac{ x^{q-1}(\ln ... 1$$1 + \sum_{k=1}^{\infty} \frac{(it)^k}{k!}a^{2k+1} = 1 - a + a\sum_{k=0}^{\infty} \frac{(it)^k}{k!}a^{2k} \\ = 1 - a + a\exp (ita^2) 

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