I am working on the following power series. (Cant copy images yet)

enter image description here

I have applied the ratio test on the series ignoring the $-ln(2)$, and have reduced it down to

$\frac{xn }{2(n+1)}$

When i take the limit of this i end up with $x/2$. I know the interval of convergence is abs$(x) < 2$.

Do i need to include the $-ln(2)$ in this part, or have i done something wrong in the calculation, as $x/2$ doesnt seem to be right. Also, does the $-ln(2)$ affect the actual values of the interval and radius of convergence.

I also concluded it was absolutely convergent for all values, and conditionally convergent for none. Is this correct?


  • 2
    $\begingroup$ Posting images of text is not appropriate on this site. Type out the question. It makes it easier to find with search engines. $\endgroup$
    – 5xum
    Sep 26, 2016 at 13:12
  • $\begingroup$ It is essential that you remember that convergence of any series is determined by the tail-end of the series. The first few terms, even a million of them, don’t affect convergence. $\endgroup$
    – Lubin
    Sep 26, 2016 at 17:57

1 Answer 1


Your power series convergent for $|x| < 2$, this is correct. You do not need to include the term $-\ln(2)$ in your considerations, because it doesn't affect the convergence or the radius of convergence. But it seems to me as if you didn't fully understand how to use the ratio test....your limit is correct, it is $\frac{|x|}{2}$ and this value has to be $< 1$, hence $|x| < 2$ as stated above. So don't forget the absolute value with $x$. Moreover, the power series is absolutely convergent for $|x| < 2$ and divergent for $|x| > 2$. The convergence behaviour at the boundary values $x = \pm 2$ needs to be evaluted separately. To check this, insert these two values in your power series and use appropriate convergence criteria to see if you get convergent/divergent series (Hint: [alternating] harmonic series!).

  • $\begingroup$ Thanks for the reply. Substituting x = +/- 2 gives the positive and negative harmonic series 1/n and -1/n, so it is divergent at both endpoints making the interval -2<x<2? $\endgroup$ Sep 27, 2016 at 9:33
  • $\begingroup$ Almost ;-) Substituting $x = 2$ gives the common harmonic series, which is of course divergent; hence, the point $x = 2$ belongs to the "divergence interval". Otherwise, when inserting $x = -2$, you'll end up with the alternating harmonic series, which is convergent by the Leibniz criterion (see en.wikipedia.org/wiki/Alternating_series_test), since: $(-2)^n = (-1)^n \cdot 2^n$, and the factor $2^n$ cancels out with the same factor in the denominator, leaving the series $\sum \limits_{n = 1}^\infty \frac{(-1)^n}{n}$. All in all, the interval of convergence is $[-2, 2)$. $\endgroup$
    – ComplexF
    Sep 27, 2016 at 9:49

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