ak=limn→∞m=kn∑m=11ne−m22n2
Find limk→∞ak
I tried using integral test and it resulted to nothing . ultimately I took the help of Central limit Theorem , even though it is completely wrong .
ak=limn→∞m=kn∑m=11(√2π)n√2πe−(m−0)22n2
⇒ak=limn→∞√2π∫m=knm=11n√2πe−(m−0)22n2,where m is N(0,n2)
⇒ak=limn→∞√2πP(1≤m≤kn)=limn→∞√2πP(1n≤mn≤knn)
⇒ak=√2πlimn→∞(Φ(k)−Φ(1n))=√2π(Φ(k)−12)
∴limk→∞ak=√π2
Any helpful insights?
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