Real Analysis Live - Problem Solving - Series and Convergent Criteria

Q: How does one prove convergence for $\sum_{n=1}^\infty \frac{n!}{n^n}$ without relying on the first few terms?

The approach expands $n!$ as a product $n(n-1)\cdots 1$ and compares factors against $n^n$. By grouping terms, the solution shows that for sufficiently large $n$, the term $a_n=\frac{n!}{n^n}$ is bounded above by a constant multiple of $1/n^2$ (the session arrives at an inequality of the form $a_n\le \frac{2}{n^2}$ for $n$ beyond a threshold). Since $\sum 1/n^2$ converges, the comparison test forces $\sum a_n$ to converge. Positivity of $a_n$ also means absolute convergence follows automatically.

Q: Why does partial fraction decomposition make $\sum_{n=1}^\infty \frac{1}{n(n+1)}$ computable?

Decomposing gives $\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$. When forming partial sums $S_N=\sum_{n=1}^N\left(\frac{1}{n}-\frac{1}{n+1}\right)$, almost every term cancels: $\frac{1}{2}$ appears once with a plus sign and once with a minus sign, and similarly for $\frac{1}{3},\frac{1}{4},\dots$. The remaining terms are $S_N=1-\frac{1}{N+1}$, so the limit is $\lim_{N\to\infty}S_N=1$.

Q: What’s the practical difference between convergence and absolute convergence in these exercises?

Absolute convergence means $\sum |a_n|$ converges; it’s stronger than ordinary convergence. The session notes that comparison-test statements often require absolute values because inequalities like $|a_n|\le b_n$ preserve order. In the main factorial example, all terms are positive, so $|a_n|=a_n$, making absolute convergence automatic. The discussion also reminds that a series can converge conditionally (converge but not absolutely) when positive and negative parts cancel—though that phenomenon isn’t needed for the positive-term examples here.

TL;DR

Use the comparison test by bounding a complicated term $a_{n}$ above (for convergence) or below (for divergence) with a simpler series whose behavior is known.

Briefing Cornell Notes

Briefing

A live real-analysis problem session zeroed in on two core skills for series: proving convergence/divergence with comparison and then, when possible, computing an exact sum using partial fractions and telescoping. The most important takeaway came early from the series

$n = 1 \sum \infty \frac{n !}{n ^{n}} .$ Rather than guessing from the first few terms, the solution builds a rigorous upper bound by comparing the factorial ratio to a multiple of $1/ n^{2}$ . The key move is to expand $n!$ as a product $n (n - 1) \dots 1$ and compare factor-by-factor with $n^{n}$ , showing that for sufficiently large $n$ the term $a_{n} = \frac{n !}{n ^{n}}$ is bounded by something like $\frac{2}{n ^{2}}$ . Since $\sum 1/ n^{2}$ converges, the comparison test forces $\sum a_{n}$ to converge (and because all terms are positive, absolute convergence follows automatically).

The session then pivoted to a second series designed to be computable:

$n = 1 \sum \infty \frac{1}{n ( n + 1 )} .$ Here the value is not just “convergent,” but exactly determined. The method starts with partial fraction decomposition: $\frac{1}{n ( n + 1 )} = \frac{1}{n} - \frac{1}{n + 1}$ . Plugging this into partial sums produces a telescoping cancellation: most intermediate terms cancel, leaving $S_{N} = 1 - \frac{1}{N + 1}$ . Taking $N \to \infty$ yields the exact sum $\sum_{n = 1}^{\infty} \frac{1}{n ( n + 1 )} = 1$ . A related exercise reinforced the same theme: once the series is rewritten into a telescoping form, convergence and the limit become straightforward.

A third example pushed in the opposite direction—divergence. The series

$n = 2 \sum \infty \frac{n + 5}{n ^{2} - 2 n + 1}$ was analyzed by asymptotic behavior: factoring the highest powers shows the term behaves like a constant multiple of $1/ n$ for large $n$ . Since $\sum 1/ n$ is the harmonic series and diverges, the solution uses a lower bound of the form $a_{n} \geq \frac{1}{n}$ (for all sufficiently large $n$ ) and applies the comparison test to conclude divergence.

Finally, a more advanced “compute the limit” exercise used partial fractions again, but with a different rational function. After decomposing into two simpler fractions with linear factors, the partial sums telescoped, producing an exact limit of $1/2$ . Throughout, the session emphasized practical exam strategy: first identify whether the terms resemble $1/ n^{2}$ (likely convergent) or $1/ n$ (likely divergent), then choose the right tool—comparison for convergence, and partial fractions plus telescoping when an exact value is attainable.

Cornell Notes

The session trains three linked moves for real-analysis series. First, it proves convergence of $\sum_{n = 1}^{\infty} \frac{n !}{n ^{n}}$ by bounding $a_{n}$ above with a convergent majorant like $\frac{C}{n ^{2}}$ , then applying the comparison test. Second, it computes an exact sum for $\sum_{n = 1}^{\infty} \frac{1}{n ( n + 1 )}$ by partial fraction decomposition into $\frac{1}{n} - \frac{1}{n + 1}$ , which makes partial sums telescope to $1 - \frac{1}{N + 1}$ and hence limit $1$ . Third, it proves divergence for a rational series by showing its terms dominate $\frac{1}{n}$ and comparing to the harmonic series. The overall lesson: convergence tests decide “yes/no,” while partial fractions + telescoping can deliver exact values.

How does one prove convergence for $\sum_{n = 1}^{\infty} \frac{n !}{n ^{n}}$ without relying on the first few terms?

The approach expands

n!

as a product

n (n - 1) \dots 1

and compares factors against

n^{n}

. By grouping terms, the solution shows that for sufficiently large

n

, the term

a_{n} = \frac{n !}{n ^{n}}

is bounded above by a constant multiple of

1/ n^{2}

(the session arrives at an inequality of the form

a_{n} \leq \frac{2}{n ^{2}}

for

n

beyond a threshold). Since

\sum 1/ n^{2}

converges, the comparison test forces

\sum a_{n}

to converge. Positivity of

a_{n}

also means absolute convergence follows automatically.

Why does partial fraction decomposition make $\sum_{n = 1}^{\infty} \frac{1}{n ( n + 1 )}$ computable?

Decomposing gives

\frac{1}{n ( n + 1 )} = \frac{1}{n} - \frac{1}{n + 1}

. When forming partial sums

S_{N} = \sum_{n = 1}^{N} (\frac{1}{n} - \frac{1}{n + 1})

, almost every term cancels:

\frac{1}{2}

appears once with a plus sign and once with a minus sign, and similarly for

\frac{1}{3}, \frac{1}{4}, \dots

. The remaining terms are

S_{N} = 1 - \frac{1}{N + 1}

, so the limit is

lim_{N \to \infty} S_{N} = 1

What’s the practical difference between convergence and absolute convergence in these exercises?

Absolute convergence means

\sum ∣ a_{n} ∣

converges; it’s stronger than ordinary convergence. The session notes that comparison-test statements often require absolute values because inequalities like

∣ a_{n} ∣ \leq b_{n}

preserve order. In the main factorial example, all terms are positive, so

∣ a_{n} ∣ = a_{n}

, making absolute convergence automatic. The discussion also reminds that a series can converge conditionally (converge but not absolutely) when positive and negative parts cancel—though that phenomenon isn’t needed for the positive-term examples here.

How does the divergence proof work for the rational series $\sum_{n = 2}^{\infty} \frac{n + 5}{n ^{2} - 2 n + 1}$ ?

The term is analyzed asymptotically by factoring the highest powers: the numerator grows like

n

while the denominator grows like

n^{2}

, so the fraction behaves like a constant times

1/ n

. The solution then turns this intuition into a comparison by proving a lower bound of the form

a_{n} \geq \frac{1}{n}

for all

n

beyond a threshold (here,

n \geq 2

). Since

\sum 1/ n

diverges (harmonic series), the comparison test implies the original series diverges.

When is telescoping the right tool, and what must be true for it to work?

Telescoping works when the summand can be rewritten so that consecutive terms cancel after summing—typically after partial fraction decomposition. In the examples, rewriting

\frac{1}{n ( n + 1 )}

\frac{1}{n} - \frac{1}{n + 1}

creates cancellation across the partial sums. The cancellation must occur systematically across the entire index range; otherwise, only a few terms would cancel and the limit would not simplify to a closed form.

Review Questions

For $\sum_{n = 1}^{\infty} \frac{n !}{n ^{n}}$ , what majorant series is used for comparison, and what inequality is established between $a_{n}$ and that majorant?
Show how $\frac{1}{n ( n + 1 )}$ decomposes into partial fractions and compute the resulting partial sum $S_{N}$ . What limit does it approach?
In the divergence example, what asymptotic behavior of $a_{n}$ leads to comparing it with the harmonic series?

Key Points

1
Use the comparison test by bounding a complicated term $a_{n}$ above (for convergence) or below (for divergence) with a simpler series whose behavior is known.
2
For $\sum \frac{n !}{n ^{n}}$ , factor $n!$ and compare products to show $a_{n}$ is eventually bounded by a multiple of $1/ n^{2}$ , forcing convergence.
3
Partial fraction decomposition can turn rational summands into differences like $\frac{1}{n} - \frac{1}{n + 1}$ , enabling telescoping cancellation.
4
Telescoping turns the computation of an infinite series into the limit of a simple expression for partial sums.
5
When a term behaves like $1/ n$ for large $n$ , comparing to the harmonic series is a reliable route to divergence.
6
Always read the question carefully: convergence/nonconvergence is one task; computing the exact sum requires additional structure (often telescoping).

Highlights

The factorial series ∑n=1∞​nnn!​ converges by comparison to a 1/n2-type majorant.

The identity n(n+1)1​=n1​−n+11​ makes the series telescope to an exact sum of 1.

A rational term with asymptotic size ∼1/n leads to divergence via comparison with the harmonic series.

Partial fractions plus telescoping can produce exact limits like 1/2, not just convergence.

The session repeatedly emphasized “eventually” bounds: inequalities need only hold for sufficiently large n to apply comparison tests.

Topics

Series Convergence
Comparison Test
Partial Fraction Decomposition
Telescoping Sums
Asymptotic Estimates

Mentioned

Fabian