Convergence Analysis of the Series Σ 1/n^n

The series in question is the sum of the terms 1/n^n. To determine whether this series converges or diverges, we can apply the ratio test. This test is particularly useful for series where the terms involve factorials, exponentials, or powers.

The Ratio Test Applied

First, let's determine the n-th term of the series:

an 1/n^n

Next, we calculate the ratio of consecutive terms:

rn an 1 / an (1/(n 1)^(n 1)) / (1/n^n) n^n / (n 1)^(n 1)

Now, simplifying n^n / (n 1)^(n 1) involves recognizing the behavior of the terms as n approaches infinity:

n^n / (n 1)^(n 1) (n / (n 1))^n * 1 / (n 1)

As n tends to infinity, the fraction n / (n 1) approaches 1, thus:

(n / (n 1))^n tends to e^(-1)

Therefore, the ratio can be approximated as:

rn ≈ e^(-1) / (n 1)

As n approaches infinity, the expression e^(-1) / (n 1) tends to 0. Since the limit of the ratio is less than 1, the ratio test tells us that the series converges.

Verification Using J Langauge

To further verify the convergence, we can write a function in the J programming language:

f .3 :n^~n.1i.y

This function generates the terms in the series up to n. Let's test it with a few values:

f 11f 31 0.25 0.037037f 101 0.25 0.037037 0.0039063 0.00032 2.1433e_5 1.2143e_6 5.9605e_8 2.5812e_9 1e_10

The rational fraction format for the series terms is also valuable for precision:

x:f 101 1r4 1r27 1r256 1r3125 1r46656 1r823543 1r16777216 1r387420489 1r10000000000

Summing the series using this function, we see that as n increases, the sum converges to a specific value:

/ f 11/ f 31.28703703703704/ f 101.29128599705904/ f 501.29128599706266/ f 10001.29128599706266

The series converges to approximately 1.29128599706266 as n approaches infinity.

Additional Insights from the Sequence Analysis

We can also analyze the sequence xn as the terms of the series:

xn 1 - 1/2^2 - 1/3^3 - ... - 1/n^n

This sequence is monotonically increasing, which can be shown by the fact that:

xn 1 > xn

Additionally, the sequence is bounded. To find an upper bound, we observe that:

xn ≤ 1 - 1/2^2 - 1/2^3 - ... - 1/2^n 1 - (1/2 - 1/2^n)

As n approaches infinity, this simplifies to:

xn ≤ 1 - (1/2 - 0) 3/2

By the Bolzano-Weierstrass theorem, a monotonic sequence that is bounded must converge. Therefore, the sequence xn is convergent.

Civilization advances by extending the number of important operations which we can perform without thinking of them.

Alfred North Whitehead