Undergraduate doubt about the concept of numerical Series

Question

I would like to explain my doubt with an example.

Suppose then the Series:

$$\sum^{+\infty}_{n=1} \frac{1}{n}\tag{1}$$

I KNOW that this series is divergent. But I'm struggling on WHY this series is divergent. Actually my doubt is about why we cannot always exibit a value of the Sum of the series (or the value) and why we cannot always exibit an explicit $S_{n}$ (partial value of the infinite sequence of the sum of terms of $a_{n}$) for then apply the limit $\lim_{n\to \infty} S_{n}$.

So, in order to construct such a series we need firstly a sequence, $$\{a_{n}\}^{\infty}_{n=1} =: \Bigg\{\frac{1}{n}\Bigg\}^{\infty}_{n=1} \equiv \Bigg\{\frac{1}{1},\frac{1}{2},\frac{1}{3},...\Bigg\} \tag{2} $$

Then, with this sequence we construct another sequence:

$$ \begin{cases} S_{1} = \displaystyle \sum_{k=1}^{1} a_{k} = \frac{1}{1}\\ S_{2} = \displaystyle \sum_{k=1}^{2} a_{k} = \frac{1}{1}+\frac{1}{2}\\ S_{3} = \displaystyle \sum_{k=1}^{3} a_{k} = \frac{1}{1}+\frac{1}{2}+\frac{1}{3} \\.\\.\\.\\S_{n} = \displaystyle \sum_{k=1}^{n} a_{k} = \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\\.\\.\\.\\\end{cases} $$

Called the sequence of the sum of the terms of the sequence $\Bigg\{\frac{1}{n}\Bigg\}^{\infty}_{n=1}$

$$\{S_{n}\}^{\infty}_{n=1} =: \Bigg\{ S_{1},S_{2},S_{3},...,S_{n},...\Biggr\} \equiv $$

$$ \equiv \{S_{n}\}^{\infty}_{n=1} =: \Bigg\{\Bigg(\frac{1}{1}\Bigg),\Bigg(\frac{1}{1}+\frac{1}{2}\Bigg),\Bigg(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}\Bigg),...,\Bigg(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\Bigg),...\Bigg\}$$

Then we realize that "the infinite term $(S_{\infty})$ of $\{S_{n}\}^{\infty}_{n=1}$" is the "term which contains the infinite sum of the terms of the sequence $\Bigg\{\frac{1}{n}\Bigg\}^{\infty}_{n=1}$"; this is the intuition behind the object called a series:

$$S_{\infty} \equiv \sum^{+\infty}_{n=1} \frac{1}{n} = \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+... \tag{3}$$

$$ * * * $$

So with the concepts presented above, we say that the series have a sum or a value if:

$$\sum^{+\infty}_{n=1} a_{n} = \lim_{n\to+\infty}S_{n}\equiv \lim_{n\to+\infty}\sum^{n}_{k=1} a_{n} = L \tag{4}$$

Well now becomes my doubt: It seems pretty clear and nature to me (even I knowing that is completely wrong) that with $(4)$ in mind we can make an procedure to find the sum for a series, which is:

$I)$ Get an sequence: $\Big\{\frac{1}{n}\Big\}^{\infty}_{n=1}$

$II)$ Write down the series: $\displaystyle \sum^{+\infty}_{n=1} \frac{1}{n}$

$III)$ Write the $S_{n}$: $\displaystyle S_{n} = \sum^{n}_{k=1} \frac{1}{k} = 1+...+\frac{1}{n}$

$IV)$ Take the limit: $ \displaystyle \lim_{n\to+\infty}\Big(1+...+\frac{1}{n}\Big)$

$V)$ Then by $(4)$ the value of $\displaystyle \sum^{+\infty}_{n=1}\frac{1}{n}$ is some number $\neq 0$

Well, this is a monstrosity. But I really need to understand why I cannot do what I did above. So, why sometimes we can "apply" the notion of $(4)$ and sometimes not?

Answer 1

The sum of series is the limit of $S_n$, not of $a_n$. You took the limit of $\frac{1}{n}$, while you needed to look for the limit of $S_n=1+\frac{1}{2}+...+\frac{1}{n}$. That's your mistake. Of course the sum of this series can't be $0$; this is very obvious.

Answer 2

By comparing your series with an integral you can estimate $S_n$.

Indeed on the interval $t\in[k,k+1]$ since the function $t\mapsto\frac 1t$ is decreasing you have $$\int_{k}^{k+1}\frac{dt}t\le \frac 1k\le \int_{k-1}^{k}\frac{dt}t$$

This is the result of integrating by a Riemann sum with lower and upper rectangles.

Thus you get $$\int_2^{n+1}\frac{dt}t\le \sum\limits_{k=2}^n \frac 1k\le \int_1^{n}\frac{dt}t$$

Note: we start at $k=2$ to avoid $\ln(0)$.

Which is $\ln(n+1)-\ln(2)\le S_n-1\le \ln(n)-\ln(1)$.

Thus we see that $S_n\sim\ln(n)$.

In fact there is a finer result which states that $$S_n=\ln(n)+\gamma+o(1)$$ where $\gamma\approx 0.577$ is called the Euler constant.

In particular $\lim\limits_{n\to\infty}S_n=+\infty$, and the series is divergent.

In IV) you took the limit of $a_n$ instead of the partial sum $S_n$, this is not the same.

Answer 3

Let $$S_N \equiv \sum_{n=1}^N \frac 1n $$ represent the finite partial sum.

If indeed the series converges and $$\lim_{N\to \infty}S_N =L$$

then it must be true that the difference between $S_N$ and $L$ must vanish in the limit $N \to \infty$

but $$ D_N \equiv L -S_N = \sum_{n=1}^\infty \frac 1n - \sum_{n=1}^N \frac 1n = \sum_{n=N+1}^\infty \frac 1n $$

since all the terms in the sum are positive then $D_N$ must be greater than any finite sum of terms starting at $N+1$

Consider summing the first $N$ terms of $D_N$, between $n=N+1$ and $n=2N$. we have...

$$ D_N > \sum_{n=N+1}^{2N} \frac 1n $$ the sum on the right hand side consists of $N$ terms, each of which is greater than $\frac 1{2N}$ so $$ D_N > N ( \frac 1{2N})=\frac 12 $$ This means that the sum of $N$ terms starting at $N+1$ is always greater than $\frac 12$ , independent of $N$ , So there is no value of $N$ for which $ D_N < \frac 12$ , whereas convergence requires that

$$ \lim_{N\to \infty} D_N =0 $$

Therefore the series does not converge.

Answer 4

The answer to your question is that some sequences don’t have limits, so that your fourth step “take the limit” becomes meaningless in such cases. The sequence in this case is $(S_n)$.

Answer 5

The sequence of partial sums isn't Cauchy: $\forall n\ge N\,, \mid S_{2n}-S_n\mid=\dfrac 1{2n}+\dots+\dfrac 1{n+1}\ge n\cdot \dfrac 1{2n}=\dfrac 12$.