I know it is very simple but something is going amiss.
We can see here that log(1+1n)=∫n+1ndtt
if t∈[n,n+1], 1n+1≤1t≤1n
I want to show that 1n+1≤∫n+1ndtt≤1n.
Can I straight infer that? If so what is the logic behind it? I was thinking of using Σni=1 and sandwiching ∫n+1ndtt between. But confused !
Answer
n≤t≤n+1
1n+1≤1t≤1n
∫n+1ndtn+1≤∫n+1ndtt≤∫n+1ndtn
1n+1≤∫n+1ndtt≤1n
No comments:
Post a Comment