3
$\begingroup$

$$L_2=\lim_{n→∞}\sum_{k=1}^n\frac{(k-\cos^2(k))^4}{n^5}.$$ My teacher said that when brackets at the numarator is expanded the limit of sum except $\dfrac{k^4}{n^5}$ equals $0$, so the Riemann sum becomes $\lim\limits_{n→∞}\sum\limits_{k=1}^n\dfrac{k^4}{n^5}$. I don't understand this point. Please explain this point. Sorry for my bad English.

$\endgroup$
3
$\begingroup$

$\sum_{k=1}^{n} k^{j} \leq \int_1^{n} x^{j} dx =\frac {n^{j+1} -1} {j+1}$ and $\frac {n^{j+1} -1} {n^{5}} \to 0$ as $ n \to \infty$ if $j <4$. Hence, when you expand $(k-\cos^{2}(k))^{4}$ all terms except $k^{4}$ give limit $0$. [Note that $\cos^{2} k \leq 1$].

$\endgroup$
0
$\begingroup$

$$\sum\frac{(k-\cos^2 k)^4}{n^5}=\sum \left(\frac{k^4-4k^3\cos ^2\left(k\right)+6k^2\cos ^4\left(k\right)-4k\cos ^6\left(k\right)+\cos ^8\left(k\right)}{n^5}\right)$$Every term is involving $\cos k$ except the term $k^4$. Applying summation and use algebra of limits together with the fact $\cos k$ is bounded by $1$ to get your result

$\endgroup$
0
$\begingroup$

Since $$ \sum_{k=1}^nk^3=\frac{n^2(n+1)^2}4 $$ we get $$ \begin{align} \lim_{n\to\infty}\sum_{k=1}^n\frac{\left(k-\cos^2(k)\right)^4}{n^5} &=\lim_{n\to\infty}\sum_{k=1}^n\left[\frac{k^4}{n^5}+O\!\left(\frac{k^3}{n^5}\right)\right]\\ &=\lim_{n\to\infty}\left[\sum_{k=1}^n\frac{k^4}{n^4}\frac1n+O\!\left(\frac1n\right)\right]\\ &=\int_0^1x^4\,\mathrm{d}x+0\\[6pt] &=\frac15 \end{align} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.