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$$\sum_{n=1}^\infty \left(\frac{1-n}{2+3n} \right)^n$$

The question is to determine whether the series converges or diverges. I tried to distribute the power $n$ to the numerator and the denominator then test both using $limits$. The denominator goes $\infty$ but unsure how to take care of the top using $limits$. Any suggestions?

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    $\begingroup$ look up the root test $\endgroup$ Commented Mar 28, 2016 at 23:39
  • $\begingroup$ or try the ratio test $\endgroup$ Commented Mar 28, 2016 at 23:42

1 Answer 1

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Long division tells us that $$ \frac{1-n}{2+3n} = \frac{-1} 3 \left( 1 - \frac 5 {2+3n} \right). $$

So we have $$ \sum_{n=1}^\infty \left(\frac{-1} 3\right)^n \left( 1 - \frac 5 {2+3n} \right)^n. $$

The factor $\displaystyle \left( 1 - \frac 5 {2+3n} \right)^n$ converges to a constant involving $e$ (start with $\left( 1 + \frac a n\right)^n\to e^a$ and do a bit of algebra). Hence the absolute values of the terms are bounded above by $$ C \left( \frac 1 3 \right)^n. $$

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