Mathematics › Limits › Miscellaneous
limn→∞11+n5+2425+n5+3435+n5+…+n4n5+n5=limn→∞∑r=1nr4r5+n5=limn→∞1n∑r=1nrn4rn5+1=∫01x41+x5dx=15ln1+x501=15ln2=ln25
limn→∞11+n5+2425+n5+3435+n5+…+n4n5+n5
=limn→∞∑r=1nr4r5+n5
=limn→∞1n∑r=1nrn4rn5+1=∫01x41+x5dx
=15ln1+x501=15ln2=ln25
Asked in: AP EAMCET 2022 (04 Jul Shift 1)
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