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Find the Taylor Series and Radius of Convergence for $$ f(x)= \frac1{n+x} $$ at $x=0$

Solution Attempt:

Taylor Series:

$f(0) = \frac1{n} $;

$f'(0) = -\frac1{n^{2}} ;$

$f''(0) = \frac{2}{n^{3}} ;$

$f'''(0) = \frac{-6}{n^{4}} ;$

$$ \sum_{k=0}^\infty \frac{(-1)^k}{k!} \frac{x^{k}}{n^{k+1}}.$$

Which formula should I use for radius of convergence?

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  • $\begingroup$ Do you know the ratio test? $\endgroup$ – ℋolo Apr 6 '18 at 0:32
  • $\begingroup$ Is my solution for taylor series correct? $\endgroup$ – K KA Apr 6 '18 at 0:36
  • $\begingroup$ @KKA it is not. $\endgroup$ – qbert Apr 6 '18 at 0:38
  • $\begingroup$ No, notice the constant at each derivative $\endgroup$ – ℋolo Apr 6 '18 at 0:40
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You have a mistake in your series...

$f^{(k)}(0) = \frac {(-1)^kk!}{n^{k+1}}\\ a_k = \frac 1{k!}f^{(k)}(0) = \frac {(-1)^k}{n^{k+1}}\\ f(x) = \sum_\limits{k=0}^{\infty}\frac {(-1)^kx^k}{n^{k+1}}$

But that isn't the easy way to do it.

This is much easier.

The sum of a geometric series: $\sum_\limits{k=0}^{\infty} r^k = \frac {1}{1-r}$

$f(x) = \frac {1}{n+x} = \frac 1n \left(\frac {1}{1-(-\frac {x}{n})}\right) = \frac 1n\sum_\limits{k=0}^{\infty} (\frac {-x}{n})^k = \sum_\limits{k=0}^{\infty}\frac {(-1)^kx^k}{n^{k+1}}$

Radius of convergence...

$\sum_\limits{k=0}^{n} r^k = \frac {1-r^{n+1}}{1-r}$

$\lim_\limits{n\to \infty}\frac {1-r^{n+1}}{1-r}$ converges when $|r|<1$

Similarly, $\sum_\limits{k=0}^{\infty}\frac {(-1)^kx^k}{n^{k+1}}$ converges when $|\frac {x}{n}|<1$ or $|x|<n$

Also worth noting, $f(x)$ heads to infinity as $x$ approaches $-n.$ That tends to be a solid indication of the radius of convergence.

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Observe that $$ f(x)=\frac{1}{n+x}=n^{-1}\frac{1}{1-(-x/n)}=n^{-1}\sum_{k=0}^\infty\frac{(-1)^k}{n^k}x^k=\sum_{k=0}^\infty\frac{(-1)^k}{n^{k+1}}x^k $$ for $|x/n|<1$ where we have used the geometric series $$ \frac{1}{1-x}=\sum_{m=0}^\infty x^m\quad|x|<1 $$ By uniqueness of power series in a neighbourhood of zero the result follows.

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You can use the ratio test on what you derived (although it is not the Taylor series you propose it is).

Alternatively, note that $$ \frac{1}{x+n}=\frac{1}{n}\frac{1}{x/n+1}=\frac{1}{n}\sum_{k=0}^\infty (-1)^k\left(\frac{x}{n}\right)^k= \sum_{k=0}^\infty (-1)^kx^k\left(\frac{1}{n}\right)^{k+1} $$ Where we used the geometric series $$ \sum_{k=0}^\infty y^k=\frac{1}{1-y} $$ which converges for $|y|<1$. Here, $y=-x/n$.

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