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Prove whether the following series converge or diverge.

$$\sum \limits_{n=1}^{\infty} \frac{2^n+3^n}{4^n-5^n}$$

I think this series converge so I tried to prove with ratio test and comparison test but it did not work.

Please give me ideas or hints on how to solve this question, thanks to anybody who helps.

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  • $\begingroup$ Root Test works fine, if we have to give justification. $\endgroup$ – André Nicolas Nov 29 '14 at 22:24
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$$\left|\frac{2^n+3^n}{4^n-5^n}\right|=\left(\frac35\right)^n\frac{\left(\frac23\right)^n+1}{\left|\left(\frac45\right)^n-1\right|}\le5\left(\frac35\right)^n\left[\left(\frac23\right)^n+1\right]\le10\left(\frac35\right)^n$$

Further hint: since $\;0<\left(\frac45\right)^n<\frac45\;$ , we have that

$$-1<\left(\frac45\right)^n-1<\frac45-1=-\frac15$$

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$$\dfrac{2^n+3^n}{4^n-5^n} = \dfrac{\left(\frac{2}{5}\right)^n + \left(\frac{3}{5}\right)^n}{\left(\frac{4}{5}\right)^n-1} \sim -\left(\frac{2}{5}\right)^n - \left(\frac{3}{5}\right)^n$$

when $n\to +\infty$

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  • $\begingroup$ How "when $\;n\to\infty\;$", yet you got an expression that depends on $\;n\;$ ? $\endgroup$ – Timbuc Nov 29 '14 at 22:22
  • $\begingroup$ @Timbuc $f(n)\sim g(n)$ when $n\to +\infty$ means $\lim_{n\to +\infty}\dfrac{f(n)}{g(n)} = 1$ $\endgroup$ – Petite Etincelle Nov 29 '14 at 22:24
  • $\begingroup$ @Timbuc The denominator approaches -1. The numerator was left alone. $\endgroup$ – user170231 Nov 29 '14 at 22:25
  • $\begingroup$ That I know, @Petite...yet your rightmost expression still has $\;n\;$ in it..! $\endgroup$ – Timbuc Nov 29 '14 at 22:25
  • $\begingroup$ @Timbuc In my previous comment, $g(n)$ has $n$ in it, so what's the problem? $\endgroup$ – Petite Etincelle Nov 29 '14 at 22:27
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HINT: Do a limit comparison with $\sum_{n\ge 1}\left(\frac35\right)^n$. It may be helpful to note that

$$\frac{2^n+3^n}{3^n}=\left(\frac23\right)^2+1\;,$$

and $$\large\frac{5^n}{4^n-5^n}=\frac1{\frac{4^n-5^n}{5^n}}\;.$$

The motivation here is that $2^n+3^n\approx 3^n$ and $4^n-5^n\approx -5^n$ for very large $n$.

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HINT: $$ \lim_{n\to\infty}\frac{\dfrac{2^n+3^n}{5^n-4^n}}{\dfrac{3^n}{5^n}}=1. $$

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  • $\begingroup$ Sorry but can you explain a bit more, I didn't quite get it. $\endgroup$ – Lucy Nov 29 '14 at 22:27
  • $\begingroup$ @Lucy Is the short extension of my answer enough for you? $\endgroup$ – Kola B. Nov 29 '14 at 22:35
  • $\begingroup$ This actually made me slightly more confused because the ratio test I was taught in lectures seem different to what you have shown to me. $\endgroup$ – Lucy Nov 29 '14 at 22:38
  • $\begingroup$ @KolaB. I think you meant the limit comparison test. That's not the ratio test. $\endgroup$ – Timbuc Nov 29 '14 at 22:39
  • $\begingroup$ @Lucy Isn't it in the form: If, for nonnegative $a_n$, $b_n$, $\lim_n (a_n/b_n)=k$, where $0<k<\infty$, then both $\sum_n a_n$ and $\sum_n b_n$ converge (or diverge)? $\endgroup$ – Kola B. Nov 29 '14 at 22:42

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