Let an=1√n2+1+1√n2+2....1√n2+n+1
Find limn→∞an.
I know the limit is zero. I also know that I need to use the squeeze theorem to solve it.
Now using n√n2+1≤an≤n√n2+n+1 as my upper and lower bounds doesn't yield 0, it yields 1.
It doesn't seem right but can I just use 1√n2+1≤an≤1√n2+n+1 to solve ?
Thanks.
Answer
The limit is not 0, in fact you are almost there!
n+1√n2+n+1≤an≤n+1√n2+1
For LHS, limn→∞n+1√n2+n+1=limn→∞1+1n√1+1n+1n2=1.
For RHS, limn→∞n+1√n2+1=limn→∞1+1n√1+1n2=1
Hence, by Squeeze Theorem, limn→∞an=1
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