Convergence of Series: Proving the Relationship Between Two Converging Series

[pository]

Our instructor assigned a problem from Rudin's Principles of Mathematical Analysis; a problem which I have been unable to solve after giving it good thought.

The statement is:
"Prove that the convergence of SUM[a_n] implies the convergence of SUM[sqrt(a_n)/n], if a_n >= 0."

The instructor did give us a hint: "Review the ideas of chapter one," from which I gleaned the archimedean property or supremums might be important. Anyway, if anyone is familiar with how this can be proved, I would appreciate a nudge in the right direction. Thanks.

 

[pository]

In case anyone else read this, I have found the solution on the internet. It was embarassingly simple. By the AM-GM inequality:

$$a_n + \frac{1}{n^2} \ge 2\frac{\sqrt{a_n}}{n}.$$​

The left hand series converges, so by direct comparison, the right hand series also converges.

 

[quasar987]

Cool, I didn't know of this inequality. What does AM-GM stands for?

 

[benorin]

Harmonic - Geometric - Arithmetic Mean Inequality

AM-GM = Arithmetic Mean - Geometric Mean

In full, the Harmonic - Geometric - Arithmetic Mean inequality is (for sequences)

$$\frac{n}{\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdot\cdot\cdot +\frac{1}{a_{n}}}\leq \left( a_{1}a_{2}\cdot\cdot\cdot a_{n}\right) ^{\frac{1}{n}}\leq \frac{a_{1}+a_{2}\cdot\cdot\cdot + a_{n}}{n}$$

where equality holds iff the $a_{i}$'s are all equal, and it is understood that $a_{i}\geq 0,\forall i$.

 

[benorin]

Harmonic - Geometric - Arithmetic Mean inequality is (for functions)

Suppose that the real-valued function $$f(x)$$ is defined, properly integrable, and strictly positive on the interval $$\left[ x_{1}, x_{2}\right]$$.

Then the Harmonic - Geometric - Arithmetic Mean inequality is (for functions)

$$\frac{x_{1} - x_{2}}{\int_{x_{1}}^{x_{2}} \frac{dx}{f(x)}} \leq \exp\left( \frac{1}{x_{1} - x_{2}}\int_{x_{1}}^{x_{2}} \log f(x) dx\right) \leq \frac{1}{x_{1} - x_{2}} \int_{x_{1}}^{x_{2}} f(x)dx$$

 

[quasar987]

Great, I'm noting all of this down.