0

I could easily proved the following: $$\frac12<\sum_{k=1}^n \frac1{n+k}<1.$$ But, I am not able to show that the sum is actually less than $3/4.$ May I get a hint?

Usual_Learner
  • 435

1 Answers1

1
  • For the first inequality, you need $n>1$ and it's direct since you sum positive terms with the first one being $1/2$.
  • Hint for the second inequality : $$\sum_{k=1}^n \frac{1}{n+k}=\frac{1}{n}\sum_{k=1}^n \frac{1}{1+\frac{k}{n}}.$$
SacAndSac
  • 1,677
  • First of all thank you sir... As far as I've understood, the later expression resembles, for $n\to\infty$, the Riemann sum for the integral $\int_0^1 \frac1{1+x},dx$ which evaluates to $\ln2\approx 0.69<0.75=3/4.$ – Usual_Learner Apr 08 '22 at 17:15