0
$\begingroup$

I don't know how to make the following limit $$\displaystyle \lim_{n\to \infty}\frac2n\sum_{k=1}^n \left(2+\frac kn\right)\left( \ln\left(2+\frac kn \right)\right)$$

into a definite integral and just need some guiding help anything will help thanks.

$\endgroup$
  • 1
    $\begingroup$ What is your problem? Hint: $\int_0^1 f(x)\,dx = \lim_{n\to\infty}\tfrac 1 n\sum_{k=1}^nf(\tfrac k n)$. $\endgroup$ – amsmath Sep 8 '18 at 22:37
2
$\begingroup$

We have that

$$\lim_{n\to \infty}\frac1n\sum_{k=0}^{n} f\left(a+{k\over n}(b-a)\right)=\int_a^b f(x) dx$$

and therefore

$$\lim_{n\to \infty}\frac2n\sum_{k=0}^n \left(2+\frac kn\right)\left( \ln\left(2+\frac kn \right)\right)=2\int_0^1(2+x)\ln(2+x)\,dx$$

$\endgroup$

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.