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I need to show whether $\displaystyle\sum_{n=1}^{\infty}{\frac{2^n+3^n}{4^n-5^n}}$ converges or diverges using the ratio test.

So far I have $\dfrac{a_{n+1}}{a_n} = \dfrac{2^{n+1}+3^{n+1}}{4^{n+1}-5^{n+1}} . \dfrac{4^n-5^n}{2^n+3^n}$

I know I could maybe use division by $5^n$ at some point but I am not sure how I could simplify from here in order to obtain an expression that I could easily show has a limit $<1$ for convergence or $>1$ for divergence.

The only tests I can utilise at the moment are ratio and comparison, along with the use of geometric series. I would easily be able to solve this if not for the minus sign in the denominator by using comparison with geometric series for example.

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  • $\begingroup$ You don’t have the limit comparison test? $\endgroup$
    – Brian M. Scott
    Nov 30, 2014 at 19:17
  • $\begingroup$ No, or at least not currently. $\endgroup$
    – MathsUndergrad
    Nov 30, 2014 at 19:18

2 Answers 2

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HINT:

$$\begin{align*} \dfrac{a_{n+1}}{a_n} &= \dfrac{2^{n+1}+3^{n+1}}{4^{n+1}-5^{n+1}} \cdot \dfrac{4^n-5^n}{2^n+3^n}\\\\ &=\frac{3^{n+1}\left(\left(\frac23\right)^{n+1}+1\right)}{5^{n+1}\left(\left(\frac45\right)^{n+1}-1\right)}\cdot\frac{5^n\left(\left(\frac45\right)^n-1\right)}{3^n\left(\left(\frac23\right)^n+1\right)} \end{align*}$$

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  • $\begingroup$ Thanks, I have rearranged further from that by cancelling powers of n+1 with powers of n etc. but cannot see where to go next. $\endgroup$
    – MathsUndergrad
    Nov 30, 2014 at 19:46
  • $\begingroup$ @MathsUndergrad: After the cancellation, you should have $\frac35$ times something whose limit as $n\to\infty$ can easily be computed. $\endgroup$
    – Brian M. Scott
    Nov 30, 2014 at 19:53
  • $\begingroup$ Ahh yes of course, what would be the best way to show/say that something like $\frac{(\frac{2}{3})^{n+1}+1)}{(\frac{2}{3})^n+1}$ converges to 1? $\endgroup$
    – MathsUndergrad
    Nov 30, 2014 at 19:58
  • $\begingroup$ @MathsUndergrad: Just observe that numerator and denominator both converge to $1$ and use the basic fact that if $a_n\to a$, $b_n\to b$, and $b\ne 0$, then $\frac{a_n}{b_n}\to\frac{a}b$. $\endgroup$
    – Brian M. Scott
    Nov 30, 2014 at 20:05
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You can also use $\displaystyle\frac{|a_{n+1}|}{|a_n|}=\frac{2^{n+1}+3^{n+1}}{5^{n+1}-4^{n+1}} \cdot \frac{5^n-4^n}{2^n+3^n}=\frac{2^{n+1}+3^{n+1}}{2^n+3^n}\cdot\frac{5^n-4^n}{5^{n+1}-4^{n+1}}$

$=\displaystyle\frac{2(\frac{2}{3})^n+3}{(\frac{2}{3})^n+1}\cdot\frac{1-(\frac{4}{5})^n}{5-4(\frac{4}{5})^n}\to \frac{3}{1}\cdot\frac{1}{5}=\frac{3}{5}$

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