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I'm trying to find the radius of convergence for $$f(x)=\int_0^x \frac{\sin(t)}{t} dt$$

It's well known that

$$\sin(t)= t -\frac{t^3}{3!}+\frac{t^5}{5!} -\frac{t^7}{7!}+\cdots$$


$$\frac{\sin(t)}{t}= 1 -\frac{t^2}{3!}+\frac{t^4}{5!} -\frac{t^6}{7!}+\cdots$$


$$f(x)=\int_0^x \frac{\sin(t)}{t}dt=x -\frac{x^3}{3\cdot3!}+\frac{x^5}{5\cdot5!} -\frac{x^7}{7\cdot7!}+\cdots=\sum \frac{x^{2n+1}}{\left(2n+1\right)\left(\left(2n+1\right)!\right)} $$

But after finding the power series for this function I have no clue on how to find the radius of convergence. Any help would be appreciated!

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you should check the given expression for the sum. its missing the term for alternating sign. – franklin Apr 9 '13 at 1:58
up vote 3 down vote accepted

Note that each term is dominated by $x^n/n!$, and the radius of convergence of this series (which is $e^x$) is $\infty$.

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Observe that $$ \left|\frac{x^{2n+1}}{(2n+1)(2n+1)!}\right|\leqslant \frac{|x|^{2n+1}}{(2n+1)!} $$ and compare to the series of the exponential function.

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Using the method described here to calculate radius of convergence.

Let $S_K(x)$ be the series approximation of $f(x)$ and $L = \lim_{k \to \infty}|S_{k+1}(x) (S_k(x))^{-1}|$. For the series you present we get:

$$ \begin{align} L &= \lim_{k \to \infty} \bigg|\frac{(-1)^{k+1}(x^{2k+3})}{(2k+3)(2k+3)!} \frac{(1+2k)(1+2k)!}{(-1)^k(x^{2k})} \bigg|\\ &=x^3 \lim_{k \to \infty}\bigg(\frac{(1+2k)(1+2k)!}{(2k+3)(2k+3)!} \bigg) \end{align}$$

through a little manipulation we can see that this simplifies to $L = 0$. For all $L < 1$, the appropriate ratio test shows that the series is convergent. $0 < 1$ so this function is convergent everywhere. i.e. the convergent interval is $(-\infty, \infty)$.

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