I am working through the proof of the Strong Law of Large Numbers in Durrett's Probability: Theory and Examples, and I have encountered a strange inequality that I cannot wrap my head around.

Suppose $\alpha>1$ and let $m\in \mathbb{N}$ be any integer. In the proof (bottom of page 74 in the fourth edition), Durrett claims that

\begin{align*} \sum_{n: \,\alpha^{n} \geq m} \alpha^{-2n} \leq \frac{1}{m^{2}}\sum_{n=0}^{\infty} \alpha^{-2n} \end{align*}

I have tried for a long time to prove this, but I am having trouble! Can anyone provide a proof of this inequality?

  • $\begingroup$ Can you please tell me the page in this edition? services.math.duke.edu/~rtd/PTE/PTE4_1.pdf Thanks $\endgroup$ – Jimmy R. Nov 22 '16 at 19:22
  • 1
    $\begingroup$ Call $N(m)=\inf\{n\mid \alpha^n\geqslant m\}$, then $$\sum_{n: \,\alpha^{n} \geqslant m} \alpha^{-2n}=\sum_{n=N(m)}^\infty \alpha^{-2n}=\sum_{k=0}^\infty \alpha^{-2(k+N(m))}=\alpha^{-2N(m)}\sum_{k=0}^\infty \alpha^{-2k}$$ Finally, note that, by the definition of $N(m)$, $$\alpha^{-2N(m)}\leqslant\frac1{m^2}$$ $\endgroup$ – Did Nov 22 '16 at 19:24
  • 1
    $\begingroup$ @JimmyR. It is page 65 in the edition you provided $\endgroup$ – möbius Nov 22 '16 at 19:25
  • $\begingroup$ @Did Looks good, thank you $\endgroup$ – möbius Nov 22 '16 at 19:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.