Evaluate $\lim_{n\to\infty}\frac{1}{\sin(n)}-\frac{1}{n}$ and $\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{n-kx}$ using L’Hospital and Riemann sum

analysiscalculusconvergence-divergencelimitsreal-analysis

I have to calculate the two following limits:

a) $\lim_{n\to\infty}\frac{1}{\sin(n)}-\frac{1}{n}$.

b) $\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{n-kx}$ for $-1<x<1$

Hint : use L'Hospital and Riemann sums.

a) So first I get common denominator $\lim_{n\to\infty}\frac{n-\sin(n)}{n\sin(n)}$, then I use L'Hospital $\lim_{n\to\infty}\frac{1-\cos(n)}{\sin(n)+n\cos(n)}$. Now, if $n$ is not an odd multiple of $\frac{\pi}{2}$, we get $0$. If it's an odd multiple, we get $\pm 1$. Now I'm not sure about my method because WolframAlpha gets another result : https://www.wolframalpha.com/input/?i=limit+as+n+approaches+infinity+of+1%2Fsin(n)-1%2Fn. Their result is $-\infty$ to$-1$, $1$ to $\infty$

b) Here I thought about factoring out an $\frac{1}{n}$ and making the substitution $x=\frac{k}{n}$, so we get

$\int_{0}^{1}\frac{1}{1-x^2}dx=\int_{0}^{1}\frac{1}{(1-x)(1+x)}dx$. Now we could make partial fraction decomposition to get $\frac{1}{2(x+1)}-\frac{1}{2(x-1)}$. And so, if we integrate that, we get $\frac{1}{2}\log(x+1)-\frac{1}{2}\log(x-1)$. Now what I find strange is that first we can let $x=\frac{k}{n}$ if $x$ is already in the equation. Second, if we evaluate that from $0$ to $1$, we get $\frac{1}{2}\log(2)+\frac{1}{2}\log(-1)-\frac{1}{2}\log(0)$ so I don't know if my approach is correct.

Thanks for your help !

Edit : enter image description here

Edit 2: For b), as said, we need to use another variable, so we get $\int_0^1\frac{1}{1-yx}dy=-\frac{\log(1-x)}{x}$ which seems valid if, as given $-1<x<1$.
For a), as said, the limit does not exist. They probably meant the limit as n approaches 0. In this case, we can use l'Hospital a second time to get $\frac{\sin(n)}{\cos(n)+\cos(n)-n\sin(n)}$ which gives $0$ as n approaches zero.

Best Answer

  1. Technically speaking, it is illegal to use L'Hopital rule to sequential limits. And I don't think such limit exists. Since the hint is the L'Hopital rule, I think it is more likely to be $$ \lim_{x \to 0} \frac 1{\sin x} - \frac 1x. $$ To let the limit be nonzero, maybe it also could be $$ \lim_{x \to 0} \frac 1{\sin^2 x} - \frac 1{x^2}. $$
  2. You got the letters wrong. $x$ is a given constant. To write Riemann sum you should consider the function $$ f(t) = \frac 1{1 - x t}, t \in [0,1]. $$

UPDATE

If you insist, then such limit does not exist.

Proof. Assume such limit exists, let it be $A$, then using the arithmetic operation of limits, $$ \lim_{n \to \infty} \frac 1{\sin n} = \lim_{n \to \infty} \frac 1{\sin n} - \frac 1n + \lim_{n \to \infty} \frac 1n = A + 0 = A. $$ Easy to see that $A \neq 0$ because $$ \left\vert \frac 1 {\sin n} \right \vert \geqslant 1. $$ Then using the arithmetic operation again, $$ \lim_{n\to \infty} \sin n = \frac 1A $$ exists. But in fact $\sin n$ has no limits [proof omitted, if you need then I will add], contradiction. Hence the limit does not exist. $\square$

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