Let tn=1n(1+1√2+⋯+1√n), n=1,2,… then I want to know if ∑∞n=1tn converges/diverges and the sequence{tn} converges and diverges
for it I thought of finding limn→∞tn=limn→∞1n∑nr=11√r but how to solve this limit I can do it if it is presented as a Riemann sum like if there is n in the denominator of r
Answer
First tn>1/n, so ∑∞n=1=∞.
For tn alone,
tn=1nn∑k=11√k=(1nn∑k=11√k/n)1√n.
The expression in brackets converges to ∫101√tdt=2, so the product converges to zero:limn→∞tn=limn→∞(1nn∑k=11√k/n)limn→∞1√n=2×0=0.
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