Question

How do i know if the series converges?

Answer 1

\[\sum_{k=1}^{\infty} ak X^{k}, ak=\sqrt{k+1} - \sqrt{k}\]

Answer 2

can you help me out?

Answer 3

I believe you mean \(\sum\limits_{k=1}^\infty a_kx^k\) with \(a_k=\sqrt{k+1}-\sqrt k\)?

You could just use the ratio test. The series will converge for any \(x\) such that
\[\lim_{k\to\infty}\left|\frac{a_{k+1}x^{k+1}}{a_kx^k}\right|=|x|\lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|<1\]

Answer 4

yes that is what i meant..sorry for late reply i didn't have internet for couple of days

Answer 5

okay, but what do i do with \[ak=\sqrt{k+1} -\sqrt{k}\] ?

Answer 6

You use the definition of \(a_k\) to compute the limit above:
\[\lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|=\lim_{k\to\infty}\frac{\sqrt{k+2}-\sqrt{k+1}}{\sqrt{k+1}-\sqrt k}\]The current limand is problematic since it's approaching an indeterminate form \(\dfrac{\infty-\infty}{\infty-\infty}\).

One way to circumvent this is by rationalizing the denominator. To save some space, denote \(A=\sqrt{k+2}\), \(B=\sqrt{k+1}\), and \(C=\sqrt k\). Then we
\[\begin{align*}
\lim_{k\to\infty}\frac{A-B}{B-C}&=\lim_{k\to\infty}\frac{A-B}{B-C}\times\frac{B+C}{B+C}\\[1ex]
&=\lim_{k\to\infty}\frac{AB-BC+AC-B^2}{B^2-C^2}
\end{align*}\]Notice that
\[B^2-C^2=\left(\sqrt{k+1}\right)^2-\left(\sqrt k\right)^2=(k+1)-k=1\]so the denominator disappears.

Now let's split up the limit like so
\[\lim_{k\to\infty}\left(AB-BC+AC-B^2\right)=\lim_{k\to\infty}(AB-B^2)+\lim_{k\to\infty}(AC-BC)\]and consider each resulting limit separately. Rationalizing will be useful again.

For the first limand, you can rewrite to get
\[\begin{align*}
AB-B^2&=(AB-B^2)\times\frac{AB+B^2}{AB+B^2}\\[1ex]
&=\frac{(AB)^2-B^4}{AB+B^2}\\[1ex]
&=\frac{\left(\sqrt{(k+2)(k+1)}\right)^2-\left(\sqrt{k+1}\right)^4}{\sqrt{(k+2)(k+1)}+\left(\sqrt{k+1}\right)^2}\\[1ex]
&=\frac{(k+2)(k+1)-(k+1)^2}{\sqrt{(k+2)(k+1)}+(k+1)}\\[1ex]
&=\frac{(k+2)-(k+1)}{\sqrt{\dfrac{k+2}{k+1}}+1}\\[1ex]
&=\frac{1}{\sqrt{\dfrac{k+2}{k+1}}+1}\\[1ex]
&=\frac{1}{\sqrt{\dfrac{1+\frac{2}{k}}{1+\frac{1}{k}}}+1}
\end{align*}\]
For the second, you can rewrite to get
\[\begin{align*}
AC-BC&=(AC-BC)\times\frac{AC+BC}{AC+BC}\\[1ex]
&=\frac{(AC)^2-(BC)^2}{AC+BC}\\[1ex]
&=\frac{\left(\sqrt{k(k+2)}\right)^2-\left(\sqrt{k(k+1)}\right)^2}{\sqrt{k(k+2)}+\sqrt{k(k+1)}}\\[1ex]
&=\frac{k(k+2)-k(k+1)}{\sqrt{k^2}\left(\sqrt{1+\dfrac{2}{k}}+\sqrt{1+\dfrac{1}{k}}\right)}\\[1ex]
&=\frac{k((k+2)-(k+1))}{|k|\left(\sqrt{1+\dfrac{2}{k}}+\sqrt{1+\dfrac{1}{k}}\right)}\\[1ex]
&=\frac{1}{\sqrt{1+\dfrac{2}{k}}+\sqrt{1+\dfrac{1}{k}}}
\end{align*}\]where the last two step uses the fact that \(|k|=k\) for \(k\ge0\).

All this to say
\[\begin{align*}|x|\lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|&=|x|\lim_{k\to\infty}\frac{\sqrt{k+2}-\sqrt{k+1}}{\sqrt{k+1}-\sqrt k}\\[1ex]&=|x|\lim_{k\to\infty}\left(\frac{1}{\sqrt{\dfrac{1+\frac{2}{k}}{1+\frac{1}{k}}}+1}+\frac{1}{\sqrt{1+\dfrac{2}{k}}+\sqrt{1+\dfrac{1}{k}}}\right)\end{align*}\]