Let a and b be positive numbers.

Prove that the series $\sum_{k=1}^{\infty}(ln(ak+b)- ln(ak))$ diverges.

At first I thought expanding it would mean a few terms get cancelled out but it only works out for a few values of a and b. But then I realized that wouldn't work out since the sum is to infinity.

Any hints on how to approach this?

  • 2
    $\begingroup$ Since $\ln(1+x) \approx x$, we see $\ln(ak+b) - \ln(ak) = \ln\left(1+\frac{b}{ak} \right) \approx \frac{b}{ak}$, so $\sum_{k=1}^{\infty} \ln(ak+b) - \ln(ak) \approx \frac{b}{a} \sum_{k=1}^{\infty} \frac{1}{k}$, which is the harmonic series, i.e. diverges. $\endgroup$ – Sameer Kailasa Feb 27 '17 at 0:46
  • 1
    $\begingroup$ The cancellation idea can be made to work, but in a restricted setting. For example, if $b=a$ then the $n$th partial sum is $$\sum_{k=1}^n\log(ak+b)-\log(ak)=\sum_{k=1}^n\log(a(k+1))-\log(ak)=\log(a(n+1))-\log(a)$$ and the RHS goes to infinity hence the series diverges. $\endgroup$ – Did Feb 27 '17 at 0:49
  • $\begingroup$ Surely since $a$ and $b$ and numbers, you can use the limit comparison test against the case where $a=b$? In this case though, the sum telescopes with unbounded partial sums. $\endgroup$ – Alfred Yerger Feb 27 '17 at 3:33


Note that $\log(1+x)\ge \frac{x}{1+x}$. Then, we have

$$\log(ak+b)-\log(ak)=\log\left(1+\frac{b}{ak}\right)\ge \frac{b}{ak+b}$$

  • $\begingroup$ Please let me know how I can improve this answer too. As always, I really want to give you the best answer I can. If the answer was not useful, I am happy to delete it. -Mark $\endgroup$ – Mark Viola Apr 12 '17 at 17:18

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