Comparison test of infinite series

[isukatphysics69]

Homework Statement

Homework Equations

The Attempt at a Solution

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

 

[BvU]

You are not worried about the terms themselves, you are worried about the sum.

and 1/3n converges
...
plug in infinity to 1/3n and you get 0 which converges

No. $\displaystyle {\sum{1\over n}}$ does NOT converge, so 1/3n doesn't either.

 

[isukatphysics69]

I was mistaken you are right thank you

 

[isukatphysics69]

i don't know how i will solve the Riemann hypothesis if i cannot even solve this stuff

 

[BvU]

Cheer up. Insight grows. From exercises that you can do easily, you don't learn fast enough. And insight grows.
And you don't have to solve the thing, just build some understanding.

 

[Mark44]

i can use the comparison test using 1/3n and 1/3n converges

so 1/3n doesn't either

Per the rules of PEMDAS, 1/3n is normally taken to mean $\frac 1 3 n$, not $\frac 1 {3n}$. Written inline this fraction should be 1/(3n) so as to not be ambiguous.

 

[Mark44]

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.

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 $\lim_{n \to \infty} \frac 1 3 n = \frac 1 3 \lim_{n \to \infty}n = \infty$.