calculus - Prove that $sum frac{1}{n^2} = frac{pi^2}{6}$

In this answer two sequences are mentioned.
In particular, I would like to prove that

$$\sum_{n = 1}^{+ \infty} \frac{1}{n^2} = \frac{\pi^2}{6}$$

If I knew that the sequence converges to $\frac{\pi^2}{6}$, I could use the $\epsilon$-$M$ criterion to prove the convergence to that value.

But how to prove that the above sequence converges to that value if I don't know the value itself? Is there a general way to proceed in such cases?

calculus - Proving Mean Value Theorem with Rolle's Theorem?

How to prove the Mean Value Theorem using Rolle's Theorem? I am getting the impression that it is possible by adding a linear function to a function where Rolle's theorem applies to prove the MVT. However, I can't quite turn this idea into a rigorous mathematical argument.

Answer

For $f$ continuous on $[a,b]$ and differentiable on $(a,b)$, the standard proofs I've seen use the function that gives the difference of $f$ and the function whose graph is the line segment joining the points $\bigl(a,f(a)\bigr)$ and $(b,f(b)\bigr)$;
$$
\phi(x)=f(x)-f(a)-{f(b)-f(a)\over b-a}(x-a).
$$

From the continuity and differentiablity of $f$ (and standard theorems such as the difference of continuous functions is continuous) it follows that $\phi$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Since $\phi(a)=\phi(b)=0$, Rolle's theorem applies to $\phi$ on $[a,b]$. Writing down the result obtained from Rolle's Theorem gives all that is desired.

sequences and series - find $frac{3}{6}+frac{3cdot5}{6cdot9}+frac{3cdot5cdot7}{6cdot9cdot12}+cdots$

find $\frac{3}{6}+\frac{3\cdot5}{6\cdot9}+\frac{3\cdot5\cdot7}{6\cdot9\cdot12}+\cdots$

I had $(1-x)^{-\frac{p}{q}}$ in mind. $$S=\frac{3}{6}+\frac{3\cdot5}{6\cdot9}+\frac{3\cdot5\cdot7}{6\cdot9\cdot12}+\cdots$$ $$S+1=1+\frac{3}{6}+\frac{3\cdot5}{6\cdot9}+\frac{3\cdot5\cdot7}{6\cdot9\cdot12}+\cdots$$ $$S+1=1+\frac{3}{2!\cdot3}+\frac{3\cdot(3+2)}{3!\cdot9}+\frac{3\cdot(3+2)\cdot(3+4)}{4!\cdot27}+\cdots$$ $$S+1=1+\frac{3}{2!}\left(\frac{\frac{2}{3}}{2}\right)+\frac{3\cdot(3+2)}{3!}\left(\frac{\frac{2}{3}}{2}\right)^2+\frac{3\cdot(3+2)\cdot(3+4)}{4!}\left(\frac{\frac{2}{3}}{2}\right)^3+\cdots$$ $$S+1=\left(1-\frac{2}{3}\right)^\frac{-3}{2}$$ I got $S=3\sqrt{3}-1$

But answer given is $S=3\sqrt{3}-4$

Answer

In the last step, you miss some multiple of $3$, and you miss one term of the expansion.

$$ S=\sum_{n\geq 1} \frac{(2n+1)!!}{(n+1)!3^n}=\sum_{n\geq 1} \binom{-\frac{1}{2}}{n+1}\frac{(-2)^{n+1}}{3^n}=3\sum_{n\geq 1} \binom{-\frac{1}{2}}{n+1}\left(\frac{-2}{3}\right)^{n+1}=3\left(\left(1-\frac{2}{3}\right)^{-\frac{1}{2}}-1-\frac{1}{3}\right)=3\sqrt{3}-4. $$

calculus - Evaluating $sum_{n=1}^{infty}frac{n}{16^n}$

I'm trying to evaluate the sum of the following infinite series:

$$\sum_{n=1}^{\infty}\frac{n}{16^n}$$

I know it converges to $\frac{16}{225}$, but I don't know how to reach this solution. It's not a geometric series or a telescoping sum, and I haven't found any way to relate it to a Taylor or Maclaurin series. How should I approach this problem?

Answer

Hint:

What is the derivative of $\;\sum_{n=1}^{\infty} x^n$?

real analysis - Given $a_n > 0$ for all $n$ and $sum a_n$ converges. Show that if $b > frac{1}{2}$, then $sum_{n=1}^infty n^{-b} sqrt{a_n}$ converges.

I attempted the integral test, limit comparison test, ratio test, and root test.

Limit comparison test: $\lim \sup \frac{n^{-b} \sqrt{a_n}}{a_n} < \infty$?

I get $\frac{0}{0}$ and apply l'Hopital's rule. (Note: I believe that when applying l'Hopital's rule, I take the derivative with respect to $n$, in which case, I suppose I can think of $a_n$ as $f(n)$, and the $\frac{\partial}{\partial x} f(n) = f'(n) \to 0$ since $a_n \to 0$.)

In most cases, I'm left with a perpetual loop of $\frac{0}{0}$.

I'm wondering if I should instead approach this problem via a comparison test and find some $\sum b_n$ that converges such that $0 \leq \sum_{n=1}^\infty n^{-b} \sqrt{a_n} \leq \sum b_n$.

Can this be proven using one of the aforementioned tests?

Answer

Using the AM-GM Inequality we have

$$ n^{-b}\sqrt{a_n} \leq \frac12(n^{-2b}+a_n)$$

Apply the conditions to the two series on the right hand side and the series on the left converges by the comparison test.

calculus - Finding value that makes $f$ continuous

Let $$f(x)=\begin{cases}\dfrac{6}{x}+\dfrac{-5x+30}{x(x-5)}, & x \neq 0,5\\ \dfrac{1}{5}, & x= 0,5\end{cases}$$
Determine the value that makes $f$ continuous at $x=0$.

So we find the limit, and if the limit exists, then the limit is continuous, and the limit equals the functional value? I know how to mechanically solve this problem but I don't really understand what's going on and what I am doing. How are we redefining $f(0)$?

$\lim_{x \to 0} \dfrac{6x-30-5x+30}{x(x-5)}=\lim_{x \to 0} \dfrac{x}{x(x-5)}=-\dfrac{1}{5}$

$f(0)=-\dfrac{1}{5}$

But then what is the positive $\dfrac{1}{5}$ included in the problem? Why is the function continuous if it doesn't equal $\dfrac{1}{5}$?

How do you know which kind of discontinuities are at $x=0,5$? How do we know that at $0$ is removable discontinuity and at $5$ is infinite discontinuity?

Thank you.

Answer

The initial function is not continuous in $\ x=0$, because:

$$\lim_{x\to0}f(x)=-\frac{1}{5}≠\frac{1}{5}=f(0)$$

as you already proved; if instead it was defined in this way:

$$\ f(x) = \left\{

\begin{array}{l l}
\frac{6}{x}+\frac{-5x+30}{x(x-5)}, & \quad \text{$x≠0,5$}\\
-\frac{1}{5}, & \quad \text{$x=0,5$}
\end{array} \right.\ $$

then the function would be continuous in $\ x=0,$ but of course it keeps being not continuous for $\ x=5$, where there is an asymptote.

calculus - Why does this sequence not oscillate?

I'm confused on why this sequence converges to 0 rather than diverge because of an oscillating series, since the result would be negative if n is odd and positive if n is even

Edit for the downvote: I apologize that my question was simple, but I genuinely thought it through for a while and could not understand. I was hoping I could ask the smart and giving people in order to enlighten me.