This is my first question on the math stackexchange-website. This is an assignment question, but I've tried to detail my thought process as granularly as possible to show I'm not just being lazy. My goal in asking this question is to fill a gap in understanding.

I'm being asked to determine whether the series below converges or diverges using the comparison test:


I've identified the dominant term as $\dfrac1{2^n}$. Rewritten as $1 \cdot 2^{-n}$, where $a = 1$ and $r = 2$, it seems like a classic geometric series test. However, $r = 2$ means that $r > 1$ which means the series should diverge. But, sampling a few terms of the series gives:

$$\{1, \frac14, \frac1{12}, \frac1{80}\}$$

which implies the series converges.

Troubleshooting my own reasoning, I believe I've either made a mistake with:

  1. Identifying the dominant term,
  2. Identifying the dominant term as an application of geometric series,
  3. Rewriting the dominant term into geometric series form, or
  4. Incorrectly applying the geometric series test.


  1. $n+1$ just acts as a linear offset, whereas $2^n$ grows exponentially, so $2^n$ must be the dominant term in the denominator,
  2. Geometric series have the form $ar^n$, and $1 \cdot 2^{-n}$ matches this perfectly (unless there is a condition that $n$ must be positive),
  3. $\dfrac1{2^n}$ is the same thing as $1 \cdot 2^{-n}$, and
  4. $r = 2$ is greater than $1$, and so by geometric series test it diverges.

What have I done wrong?


  • $\begingroup$ your $r$ is $2^{-1}$ cause $n$ has to be positive. $\endgroup$ – mm-aops Nov 10 '13 at 21:53
  • $\begingroup$ You can check out the MathJaX basic tutorial for help on how to write $\LaTeX$ :) $\endgroup$ – Maethor Nov 10 '13 at 22:24

Your reasoning is correct, but notice that $r = 1/2$, not $r = 2$, so the series $\sum_ {n = 0}^\infty 2^{-n}$ converges. Also, regarding (3), $n = 0, 1, 2, \ldots$, so $n$ is nonnegative, not just positive. Moreover, your reasoning in step (4) is not completely correct because if something "larger" diverges, it does not mean something "smaller" also diverges. Think about it, the given series is $\leq \sum_{n = 0}^\infty 1/(n + 1)$, which is the harmonic series and it diverges, but your series converges.

  • $\begingroup$ I failed to recognize that n must be positive since it is defined from 0 to infinity, and so I didn't see that $$1 \cdot 2^{-n}$$ is actually $$1 \cdot (2^{-1})^n$$, which gives an $$r < 1$$ that makes $$b_n$$ convergent, which implies $$a_n$$ is also convergent since $$a_n < b_n$$. Thanks. $\endgroup$ – astudent Nov 10 '13 at 23:10
  • $\begingroup$ Sorry, can't fix the formatting b/c comments can only be edited for 5 mins. $\endgroup$ – astudent Nov 10 '13 at 23:16

Observe $$\sum_{n=0}^{\infty} \frac{1}{(n+1)2^n} < \sum_{n=0}^{\infty} \frac{1}{2^n}$$ so if $$\sum_{n=0}^{\infty} \frac{1}{2^n}$$ converges then so does the original series.

Recall that a geometric series is of the form:

$$ \sum_{k=0}^{\infty} ar^k$$

Now $$ \sum_{k=0}^{\infty} \frac{1}{2^k} = \sum_{k=0}^{\infty} 1(\frac{1}{2})^k $$

So $a=1$, $r=\frac{1}{2}$. Thus this series converges, which implies the original series also converges.


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