isukatphysics69 said:
So the book is saying that this series diverges, i have learned my lesson and have stopped doubting the authors of this book but i don't understand how this series diverges. ok i can use the comparison test using 1/3n and 1/3n converges, since 1/(3n+1) is less than 1/3n i don't see how it will not also converge? plug in infinity to 1/3n and you get 0 which converges
The topic of infinite series has some subtleties that cause confusion in many calculus students.
If a series ##\sum_{n = 1}^\infty a_n## to converges, it's irrelevant that the sequence of terms, ##\{a_n\}## converges. What is relevant is wheter the sequence of partial sums converges or diverges. In other words, whether ##\lim_{n \to \infty} S_n## converges. Here ##S_n = \sum_{k = 1}^n a_k = a_1 + a_2 + \dots + a_n##
The sequece of terms in a series can converge to zero, without the series itself converging. This means that most of the time, evaluating the limit ##\lim_{n \to \infty} a_n## is a wasted effort. The only time this is useful is when ##\lim_{n \to \infty} a_n \ne 0##, from which we can conclude by the Nth Term Test for Divergence, that the series definitely diverges. We cannot conclude anything at all if this same limit happens to be zero.
For example, it's well known that the harmonic series, ##\sum_{n = 1}^\infty \frac 1 n## diverges, even though ##\lim_{n \to \infty}\frac 1 n = 0##
For the series ##\sum_{n = 1}^\infty n##, since ##\lim_{n \to \infty}n = \infty \ne 0##, we
can conclude that this series diverges, by the Nth Term Test for Divergence.
isukatphysics69 said:
plug in infinity to 1/3n and you get 0 which converges
Two things here:
First, we don't "plug in infinity" since infinity is not a number we can use in any arithmetic or algebraic expression. Instead, we look at a limit.
Second, as I mentioned in a previous post, 1/3n means ##\frac 1 3 n##, so ##\lim_{n \to \infty} \frac 1 3 n = \frac 1 3 \lim_{n \to \infty}n = \infty##.