Convergence of an infinite sequence with factorials trick: test the convergence of 2^n/n!

Infinite sequence with factorials trick! We test convergence of 2^n/n!: an infinite sequence with factorials and exponentials in the sequence. The trick for this problem is not obvious, but sequences with factorials often require this sort of trick to test the convergence of an infinite sequence.

We begin by writing the first several terms of the sequence to look for a pattern, and we show each term is less than or equal to 4/n. This puts an upper bound on the terms of the sequence, and we also know that each term is positive, so the lower bound on the terms is zero.

Taking the limit of the upper bound for the sequence, we get that the limit of 4/n as n goes to infinity is zero. Finally, we apply the squeeze theorem for sequences to conclude that the limit of 2^n/n! is zero as well, so the sequence is convergent.

[arianeparadis]

Before finding this video, I did not understand why in similar problems, (2^n)/n was always kept for values greater than 1. Now it makes a lot of sense. Thank you for your help!

[xslayerm]

Bro explaining it better than my college professor

[Abc-osvm]

Bro but it ain't enough who told me that in the 128th term 2to the power of n will take over 128!(i know it won't) but sorry that this demonstration is really narrow and based on common sense rather than actual demonstration

[aidenfoley]

My algebra 2 teacher has us doing these same equations in a different context.

[renshakim]

This was really cool of u.... Amazing trick to solve