The number e is the least upper bound of the sequence at which you are looking.
The argument to prove boundedness is too long to write here.
It involves the inequality (1+a)k>1+ka for every integer k>1 and real number a>-1 but a not 0.
You can probably find the derivation on line; I have it in an old calculus textbook by Johnson & Kiokemesiter.
Paul M.
tutor
I am not sure I understand your question. If you have an expression for the general term of the sequence, then the limit of the term as n goes to infinity IS the limit of the sequence. If the sequence is an alternating sign sequence or if the general term involves a periodic function (e.g. sin), then some of the terms will be greater than the limit and some will be less, but the absolute value of any term beyond a certain point and the limit can be made as small as desired or needed...that is the definition of the limit.
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10/20/20
Ashley P.
Thank you very much for the response. Yes I Googled the proof and found the same as the one which you've stated. Just to clarify, does taking limit of the general term(say n th term) of the sequence, as n tends to infinity, always return the highest possible value for a sequence?10/20/20