I'm trying to find limn→∞(1√n2+1+1√n2+2+...+1√n2+n).
- I tried to use the squeeze theorem, failed.
- I tried to use a sequence defined recursively: an+1=an+1√(n+1)2+n+1. It is a monotone growing sequence, for every n, an>0. I also defined f(x)=1√(x+1)2+x+1. So an+1=an+f(an). But I'm stuck.
How can I calculate it?
Answer
It looks squeezable.
n√n2+n≤n∑k=11√n2+k≤n√n2+11√1+1n≤n∑k=11√n2+k≤1√1+1n2
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