If it is Arithmetic Progression then find the value of s?

The value of s =
I started this question by making an A.P as the common difference is same and got the answer that I need number of terms to proceed further but my valie for number of terms is coming in fraction that is not possible i tried many times but end up with the same value so I can't proceed further as I don't know the value of n to put in the sum of Arithmetic Progression series?

Answer

Hint:

The general term is $$\frac{1}{(5r-3)(5r+2)}$$

We use partial fraction decomposition, so:

$$\frac{1}{(5r-3)(5r+2)}=\frac{A}{5r-3}+\frac{B}{5r+2}$$

Can you find $A$ and $B$? (Hint 2: $A+B=0$)

Because $A+B=0$, can you see that between the decomposition of $$\frac{1}{(5r-3)(5r+2)}$$ and $$\frac{1}{(5(r+1)-3)(5(r+1)+2)}=\frac{1}{(5r+2)(5r+7)}$$

some terms are cancelled? Figure out which ones will not be cancelled and sum them.

This method is known as telescoping

analysis - Measure theory limit question

Let $(X, \cal{M},\mu)$ be a finite positive measure space and $f$ a $\mu$-a.e. strictly positive measurable function on $X$. If $E_n\in\mathcal{M}$, for $n=1,2,\ldots $ and $\displaystyle \lim_{n\rightarrow\infty} \int_{E_n}f d\mu=0$, prove that $\displaystyle\lim_{n\rightarrow\infty}\mu(E_n)=0$.

Answer

Since $f$ is almost everywhere strictly positive, the increasing sequence of sets $$A_n=\{x\in X:f(x)>1/n\}$$
has the property that $$\lim_{n\to\infty} \mu(A_n)=\mu(X).$$ Now $\int_E f ~d\mu0$.

So let $\epsilon>0$. Choose $n$ so that $\mu(X\backslash A_n)<\epsilon/2$. For $N$ large enough, $\int_{E_N}f~d\mu<\epsilon/(2n)$ and hence $$\mu(E_N\cap A_n)

limits - Prob. 12, Sec. 5.2, in Bartle & Sherbert's INTRO TO REAL ANALYSIS: Continuity of additive functions

Here is Prob. 12, Sec. 5.2, in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:

A function $f \colon \mathbb{R} \to \mathbb{R}$ is said to be additive if $f(x+y)= f(x) + f(y)$ for all $x, y$ in $\mathbb{R}$. Prove that if $f$ is continuous at some point $x_0$, then it is continuous at every point of $\mathbb{R}$.

My Attempt:

Let $c$ be an arbitrary real number.

First suppose that $x_0 = 0$. As $f$ is continuous at the point $0$, so given any real number $\varepsilon > 0$ we can find a real number $\delta > 0$ such that
$$\lvert f(x) \rvert = \lvert f(x) - 0 \rvert = \lvert f(x) - f(0) \rvert < \varepsilon $$
for all $x \in \mathbb{R}$ for which
$$ \lvert x \rvert = \lvert x-0 \rvert < \delta. $$
Then

$$ \lvert f(x) - f(c) \rvert = \lvert f( x - c ) \rvert < \varepsilon $$
for all $x \in \mathbb{R}$ such that
$$ \lvert x-c \rvert < \delta. $$
Thus it follows that $f$ is continuous at every point $c \in \mathbb{R}$.

Am I right?

Now let us suppose that $x_0 \neq 0$. As $f$ is continuous at $x_0$, so, for every real number $\varepsilon > 0$ we can find a real number $\delta > 0$ such that

$$ \left\lvert f \left(x - x_0 \right) \right\rvert = \left\lvert f(x) - f \left( x_0 \right) \right\rvert <
\varepsilon $$
for all real numbers $x$ which satisfy
$$ \left\lvert x- x_0 \right\rvert < \delta. $$

Now if we could show from here that $f$ is continuous at $0$, then as before we will have managed to show that $f$ is continuous at every real number $c$.

What next? How to proceed from here to show that our function $f$ is continuous at $0$? Or, is there any other route?

Answer

You are almost there. Let $| t-0 | = |t| < \delta$, where $\delta$ is the one you have in the second box. Then $t + x_0 = x\in \mathbb{R}$, so $|t| = |x - x_0| < \delta$.

Thus, $|f(t) - f(0)| = |f(t)| = |f(x-x_0)| = |f(x) - f(x_0)| < \epsilon$ for $|t| < \delta$.

So $f$ is continuous at $0$.

elementary number theory - Solutions to Linear Diophantine equation $15x+21y=261$

Question

How many positive solutions are there to $15x+21y=261$?

What I got so far
$\gcd(15,21) = 3$ and $3|261$
So we can divide through by the gcd and get:
$5x+7y=87$

And I'm not really sure where to go from this point. In particular, I need to know how to tell how many solutions there are.

Answer

Apply the Euclid-Wallis Algorithm,
$$
\begin{array}{r}
&&1&2&2\\

\hline\\
1&0&1&-2&5\\
0&1&-1&3&-7\\
21&15&6&3&0
\end{array}
$$
The second to last column says that $3\cdot15+(-2)\cdot21=3$, multiplying this by $87$, yields $261\cdot15+(-174)\cdot21=261$. The last column says tha the general solution is $(261-7k)\cdot15+(-174+5k)\cdot21=261$. Using $k=35,36,37$, we get the $3$ non-negative solutions:
$$
\begin{align}
16\cdot15+1\cdot21&=261\\

9\cdot15+6\cdot21&=261\\
2\cdot15+11\cdot21&=261\\
\end{align}
$$

sequences and series - Value of this sum?

Let $S_k$, $k=1,2,3,4...,100$ denote the sum of infinite geometric series whose first term is $\frac{k-1}{k!}$ and the common ratio is $\frac{1}{k}$. Then the value of $$\frac{100^2}{100!}+\sum^{100}_{k=1}|(k^2-3k+1)S_k| \text{ is?}$$

All I could figure out in this one is:

$S_k=\frac{(k-1)k}{k!(k-1)}$

But I'm not sure on how to continue. Please help

Answer

What you found for $S_k$ simplifies to $S_k={1\over(k-1)!}$. Now $k^2-3k+1\gt0$ if $k\ge3$, so you are asking for the value of

$${100^2\over100!}+S_1+S_2+\sum_{k=3}^{100}{k^2-3k+1\over(k-1)!}={100\over99!}+1+1+\sum_{k=3}^{100}{(k-1)^2-k\over(k-1)!}$$

But

$$\sum_{k=3}^{100}{(k-1)^2-k\over(k-1)!}=\sum_{k=3}^{100}{k-1\over(k-2)!}-\sum_{k=3}^{100}{k\over(k-1)!}=\sum_{k=2}^{99}{k\over(k-1)!}-\sum_{k=3}^{100}{k\over(k-1)!}={2\over1}-{100\over99!}$$

and thus

$${100^2\over100!}+\sum_{k=1}^{100}|(k^2-3k+1)S_k|=4$$

calculus - Solving the Riemann Sum $sum_{i=1}^{n}(1+frac{6i}{n})^3(frac{2}{n})$?

So I have the Riemann sum. $\sum_{i=1}^{n}(1+\frac{6i}{n})^3(\frac{2}{n})$. From my understanding that turns into $(\frac{2}{n})\sum_{i=1}^{n}(1+\frac{6i}{n})^3$ and what is really perplexing me is that that summation turns into
$$
\left(\frac{2}{n}\right)\sum_{i=1}^{n}\left(1+\frac{6i}{n}\right)^3\rightarrow \left(\frac{2}{n^4}\right)\sum_{i=1}^{n}(n+6i)^3$$From there on out I can solve the summation, but I'm having trouble understanding how the $n$ was pulled out and the new summation was formed. If someone could explain that bit I would greatly appreciate it, thanks in advance!

Answer

$$\frac{2}{n}\sum_{i=1}^{n}\left(1+\frac{6i}{n}\right)^{3}=\frac{n^{3}}{n^{3}}\frac{2}{n}\sum_{i=1}^{n}\left(1+\frac{6i}{n}\right)^{3}=\frac{2}{n\cdot n^{3}}\sum_{i=1}^{n}n^{3}\left(1+\frac{6i}{n}\right)^{3}$$

and $$n^{3}\left(1+\frac{6i}{n}\right)^{3}=\left(n+\frac{6in}{n}\right)^{3}=\left(n+6i\right)^{3}$$

sequences and series - Testing for absolute convergence? $frac{(-1)^n}{5n+1}$

I'm trying to test the summation $\sum^\infty_{n=0}\frac{(-1)^n}{5n+1}$ for absolute convergence.

By alternating series test, I can tell is is at least conditionally convergent.

However, when I used the ratio test, I got 1, which means it doesn't tell us anything.

A google search showed an answer on yahoo answers using the limit comparison test, using the harmonic series to compare it with, but they seemed to ignore the $(-1)^n$...

The answer in the book says it is conditionally convergent, but I can't work out how to show that it is not absolutely convergent.

Any ideas?