Rinkoo Gupta
Last Activity: 10 Years ago
lim n->infinity n^1/n
let A=lim n->infinity n^1/n
taking log on both sides
log A=lim n->infinity (logn)/n form infinity/infinity
Using L'Hospital rule
logA=lim n->infinity (1/n)/1
=>logA=limn->infinity(1/n)
=>logA=1/infinity=0
=>A=e^0=1
Hence gthe largest Term in this sequence is 1.
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Rinkoo Gupta
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