Can anyone help solving this?

Won't each term tend to 0. Is 1/(infinity^2)^1/2 indeterminate

I understand that the number of terms won't be finite.

@Jasmine even in $\sum_{k=1}^{n} \frac{1}{n}$ each term goes zero, but overall there is a finite sum, ie 1 as n goes infinity

@samjoe how did you say 1?

In fact use this similar series to sandwich. Required sum is $\sum_{k=0}^{2n} \frac{1}{\sqrt{n^2+k}}$. Each term can be bounded as $\frac{1}{\sqrt{n^2+2n}} \le \frac{1}{\sqrt{n^2+k}} \le \frac{1}{\sqrt{n^2}}$. Apply sum on all sides, you'll get $2 \le S \le 2$

@samjoe each term won't tend to 0 with same magnitude I guess

@Jasmine In infinite series first calculate the sum. For infinite terms, limit of sum is not sum of limits.

@samjoe you mean limit doesn't follow distributive property in case of infinite terms? Why so?

yeah

I think I understand this

Can I say since in infinite terms all terms would go on increasing increasing and increasing but by less amount so some can argue it tends to 0 while others may say it tends to Infinity?

Yeah, like you can have two examples, $\sum_{k=0}^{n} \frac{k}{n^2}$ where series converges and $\sum_{k=0}^{n} \frac{k}{n}$ where it doesnt

@samjoe I don't know what you mean by convergent and divergent I just googled it and founfmd that in convergent each term must tend to 0

Why must each term go to zero, we do have gp which do converge, that is give a finite value as number of terms increases indefinitely: $S = \lim_{n\to\infty} \sum_{k=0}^{n} 2^{-k} = 2$ etc

google.co.in/…

Google says that

Sorry I can't find it

Actually I think you mean this: the first of the tests, $\lim_{n\to \infty }{a_n} \neq 0$ implies series diverges wikipedia

Do you agree @samjoe ?

I don't know what exactly they mean by last line

@samjoe how did you calculatev sum as 2

@Jasmine I think what it means is that terms become smaller and smaller in magnitude as $n$ increases. Read further on wikipedia they give a counterexample of their statement: $\sum_{k=1}^{n}\frac{1}{k}$ diverges

What they mean is that $\lim_{n\to\infty} a_n = 0$ is not enough for convergence.

@Jasmine use sandwich theorem. You need to find such bounds for each term so that the sum of both bounds gives same limit

@samjoe then how did you get value of the sum of each bound to be 2

$\sum_{k=1}^{2n}\frac{1}{\sqrt{n^2}} = \frac{2n}{\sqrt{n^2}}$ and now apply limit. Same for lower bound

@samjoe how did you write this? I am not getting.

@Jasmine $n$ is constant for the sum, essentially you are summing same term again and again. $2n$ times.

@samjoe oh yeah n is a constant,,, I forgot that

n is an unknown constant tending to infinity

Got everything except for the convergent part thanks .

But I will go through the Wikipedia once as the topic is new to me

Actually I am just beginning to learn this topic, I don't know much about it..

@samjoe it means it's not required for jee?

Yeah i don't think its in jee.

Oh then am not going to waste time.

:P

@Abcd Here $x$ is not varying but $n$ is. Its easier to use $u = 1/n$ then limit becomes $u\to 0^+ \frac{x^u-x^{\tfrac{u}{u+1}}}{u^2} = \frac{x}{x^{\tfrac{1}{u+1}}}\cdot \color{red}{\frac{x^{\tfrac{u^2}{u+1}}-1}{u^2}} \cdot \frac{\color{red}{u+1}}{u+1}$. Limit is split on product now, since each limit exists. For red part, use $\lim_{x\to 0} \frac{a^x-1}{x} = \ln(a)$. Final answer is $\ln(x)$ because $\lim_{u\to 0} \frac{x^{\tfrac{u^2}{u+1}}-1}{\tfrac{u^2}{u+1}} = \ln(x)$.

Thanks @samjoe

$$\lim_{x \to 0 }\dfrac{\cos(\tan x) - \cos x}{x^4}$$

I used a trick since x is not varying you can use x=2 I used 2 as we are most familiar with 2 do you know that that as you go on increasing n say n=10,100,1000,10000 I used these numbers as we know that (2)^100 and (2)^101 will have the same of digits up to 3 decimal places as number of digit in 100 is 3. And ultimately you get the difference approaching ln2 i.e 0.693

To understand the last line you can check up the calculator for value of (2)^1/100 and (2)^1/101

Oh no big typo please replace my 100 to 1/100

nice @jasmine but did you do it without calculator

@samjoe How to do that question

Usinf taylor expansions gives 0 as the answer

But its not in any option

@Abcd Either do LH four times, or maybe one or two less, or use taylor. Both require patience, just collect all $x^4$ terms from numerator, lower power terms do cancel

@samjoe I used Taylor, please see what I get:

@samjoe I think I have learned the values in case of 2 so that when my brain fails to work in exam I can use tricks it will be lengthy but will give me marks

@Jasmine oh yeah that I too have learned, $\ln 2 = 0.693$, its used a lot in physics/ chemistry..

@Abcd I am getting $-1/3$

@samjoe yeah 2 my favourite numbwr

@samjoe See the taylor expansion^

@Abcd probably you missed the $x^4$ term from $-\frac{\tan^2(x)}{2!}$. It will be there, $-\frac{(x+\tfrac{x^3}{3} ...)^2}{2!}$

From there you get $-1/3$ , others cancel, only this remains.

Done thanks @samjoe

@Abcd when where you taught Newton Leibniz theorem

Our teacher started with continuity will he teach it later

Can anyone help me clear doubts of polynomial function

@Jasmine you add the question here, those who come will see it and may answer it

Ok first of all what do you mean by degree of a polynomial I mean in (x^5-1)/(1-x) what is the degree?

Can I claim the degree of rational polynomials to be infinity as they can represent infinite some of a GP

A polynomial function is by definition a function with integer as degrees

Degree is the highest power of x