Exercise 3.7

Question

Exercise 7:
    Prove that the convergence of $\sum a_n$ implies the convergence of $$\sum\frac{\sqrt{a_n}}{n},$$ if $a_n\ge 0$.

ghostofgarborg · Accepted Answer

Assume $\sum a_n = C$. We know that $\sum \frac{1}{n^2}$ converges. Let $D$ be the number it converges to. 
    Note that $ S_n = \sum_{i=0}^n \sqrt{a_i} / i$ is monotonic.
    By the Cauchy-Schwarz inequality 
    $$ \sum_{i=1}^{n} \frac{\sqrt{a_i}}{i} \leq \left( \sum_{i=1}^{n} a_i \right)^{\frac 1 2} \left( \sum_{i=0}^{n} \frac{1}{i^2} \right)^{\frac 1 2} < \sqrt{C \cdot D} $$
    This implies that $\sum \frac{\sqrt{a_n}}{n}$ is bounded, hence converges.

Amit Awasthi · Answer

\begin{proof}
      Consider the terms of the sequence $\left\{\left(\sqrt{a_n} - 1/night)^2ight\}$:
      \begin{align*}
        \left(\sqrt{a_n} - \frac{1}{n}ight)^2 &\ge 0\
        a_n - 2 \frac{\sqrt{a_n}}{n} + \frac{1}{n^2} &\ge 0\
        a_n + \frac{1}{n^2} &\ge 2\frac{\sqrt{a_n}}{n}\
        \frac{a_n}{2} + \frac{1}{2n^2} &\ge \frac{\sqrt{a_n}}{n}.
      \end{align*}
      Since the sum of two convergent series is convergent, $\Sigma \sqrt{a_n}/n$ converges via the comparison test.
    \end{proof}

Round · Answer

By the Cauchy Theorem (Theorem 3.27), it suffices to prove that $\sum_k 2^k\cdot \frac{\sqrt{a_{2^k}}}{2^k}=\sum_k \sqrt{a_{2^k}}$ converges.

Since $\sum_n a_n$ converges, $\sum_k 2^ka_{2^k}$ converges. By the root test, 
$$\limsup_{k\rightarrow \infty} (2^k a_{2^k})^{\frac{1}{k}}\leq 1,$$ 
implying 
$$\limsup_{k\rightarrow \infty} (a_{2^k})^{\frac{1}{k}}\leq \frac{1}{2},$$
implying 
$$\limsup_{k\rightarrow \infty} \sqrt{a_{2^k}}^{\frac{1}{k}}\leq \frac{1}{\sqrt{2}}<1.$$
By the root test, $\sum_k \sqrt{a_{2^k}}$ converges. Q.E.D.

Exercise 3.7

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Exercise 3.7

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