Radius of Convergence for an infinit sum

Given the following infinite sum,

∞∑n=1xn√n+1\sum _{ n=1 }^{ \infty }{ \frac { x^n }{\sqrt{n+1}} }

a) Find the radius of convergence? Find for what values of ρ\rho it is convergent and divergent?

b) Find the sum for x=−1x=-1 with an error of highest 0.010.01, by using the sum of a finite sum with of a suitable number of variables.

My Approach

So what I have done so far is the following,

a) I used the ratio to determine the radius of convergence,


|xn+1√(n+1)+1xn√n+1||\frac{\frac { x^{n+1} }{\sqrt{(n+1)+1}} }{\frac { x^n }{\sqrt{n+1}} }|

|√n+1√n+2×x||\frac{\sqrt{n+1}}{\sqrt{n+2}} \times x|

As n goes towards infinity, we get the following,


Hence therefore the radius of convergence, ρ=1\rho=1.

Now if

x=ρ=1x=\rho=1, then the sum is divergent because it is not less than 1.

x=−ρ=1x=-\rho=1, then the sum is convergent because it is less than 1.

(Is it correct how I have solve part a)?

b) For this part, I am not sure even how to start. Because how am I suppose to determine a finite sum to find the sum for the error 0.01. Any hint or help will be great.



2 Answers


Hint for (b):

\sum^\infty_{n=1} \frac{(-1)^n}{\sqrt{n}}=\sum^N_{n=1} \frac{(-1)^n}{\sqrt{n}}+\sum^\infty_{n=N+1} \frac{(-1)^n}{\sqrt{n}}
which means
|∞∑n=1(−1)n√n−N∑n=1(−1)n√n|= |1√N+1−1√N+2+1√N+3−1√N+4+…|= |√N+2−√N+1√(N+2)(N+1)|+|√N+3−√N+2√(N+3)(N+2)|+…≤ 1√N+2(N+1)+1√N+3(N+2)+…≤ ∫∞Ndx(x+1)√x+2≤∫∞Ndxx√x.\begin{align}
\left|\sum^\infty_{n=1} \frac{(-1)^n}{\sqrt{n}}-\sum^N_{n=1} \frac{(-1)^n}{\sqrt{n}} \right| =&\ \left| \frac{1}{\sqrt{N+1}}-\frac{1}{\sqrt{N+2}}+\frac{1}{\sqrt{N+3}}-\frac{1}{\sqrt{N+4}}+\ldots \right|\\
=&\ \left|\frac{\sqrt{N+2}-\sqrt{N+1}}{\sqrt{(N+2)(N+1)}} \right|+\left|\frac{\sqrt{N+3}-\sqrt{N+2}}{\sqrt{(N+3)(N+2)}} \right|+\ldots\\
\leq&\ \frac{1}{\sqrt{N+2}(N+1)}+\frac{1}{\sqrt{N+3}(N+2)}+\ldots \\
\leq&\ \int^\infty_{N} \frac{dx}{(x+1)\sqrt{x+2}} \leq \int^\infty_{N} \frac{dx}{x\sqrt{x}}.

Evaluate the last integral to get a bound on the error. Then see for what NN you will have an error of at most .01.01.

Edit: I have used the fact that
√N+2−√N+1= √N+2+√N+1√N+2+√N+1[√N+2−√N+1]= 1√N+2+√N+1≤1√N+1.\begin{align}
\sqrt{N+2}-\sqrt{N+1} =&\ \frac{\sqrt{N+2}+\sqrt{N+1}}{\sqrt{N+2}+\sqrt{N+1}}[\sqrt{N+2}-\sqrt{N+1}] \\
=&\ \frac{1}{\sqrt{N+2}+\sqrt{N+1}} \leq \frac{1}{\sqrt{N+1}}.



Thank You. So that means I have to solve (integral)=0.01 and then solve for N, right?
– captainavenger96
Oct 20 at 20:07



@captainavenger96 exactly – you solve the integral in terms of NN and ∞\infty (which should disappear) and you’ll be left with a function in terms of NN which is equal to 0.010.01, which you can then solve 🙂
– Skeleton Bow
Oct 20 at 20:10



Great. Thank you once again for your help 🙂
– captainavenger96
Oct 20 at 20:11



I have used the fact that √n+1−√n≤1√n\sqrt{n+1}-\sqrt{n} \leq \frac{1}{\sqrt{n}}.
– Jacky Chong
Oct 20 at 21:48



Which follows from √n+1+√n√n+1+√n(√n+1−√n)\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}(\sqrt{n+1}-\sqrt{n}).
– Jacky Chong
Oct 20 at 21:49

You’re almost there for a.

Yes, the radius of convergence is 11. You could also use the Cauchy-Hadamard formula, as told to me by a great mathematician by the name Don Antonio.

When x=1x=1, the sum becomes


This diverges according to the Limit Comparison Test with 1n0.5\frac{1}{n^{0.5}}.

When x=−1x=-1, it is


This converges according to the Alternating Series Test, as it a) alternates and b) is decreasing (try looking at the derivative of f(n)=|un|f(n)=\left|{u_n}\right| and see if the function is increasing or decreasing!)

I don’t know about b, sorry!



Thank You for your time and help!! 🙂
– captainavenger96
Oct 20 at 20:08



No problem! I’m glad I could be of help. Also if you choose an answer to this question, please choose the other guy, he did most of the helping 😉
– Skeleton Bow
Oct 20 at 20:10