Converge or Diverge help?

Write ln(n+1) - ln(n) = ln(1 + 1/n) to see a_n --> 0 as n --> infinity.

The sequence a_n has a divergent sum. To see this, apply the comparison test with the series b_n = 1/(2n), which diverges.

To show a_n >= b_n, define the function f(x) = ln(1 + x) - x/2.

f(0) = 0 and you can show that f'(x) > 0 for 0 <= x < 2. By the mean value theorem, this implies a_n > b_n, since f(1/n) = a_n - b_n.

a set converges if it has a decrease. If it would not have a decrease, it diverges. there isn't any single attempt this is going to in the present day inform you what takes place for any given series. you're able to take issues on a case via case foundation. consequently, one attempt we are able to apply the assessment attempt. If we are able to detect a converging series the place each term is greater suitable than the corresponding term of our series, then our series could desire to converge too. i be attentive to (a million + (a million/n) )^n converges to e (2.71828...), via fact is the definition of e. for the reason that a million/n > a million/2n > 0 for n >=a million, this suggests a million+a million/n > a million + 2/n > a million, so (a million + a million/n)^n > (a million + a million/2n)^n. So the series converges. On a part word, it would not converge to a million, as somebody else pronounced. It easily converges to sqrt(e).