# Convergence of the series $\frac1{n^n}$

Does the series $\frac1{n^n}$ converges or diverges .

By comparison test I can claim that $\frac1{n^n} < \frac1n$ , and since series $\frac1n$ appears divergent , so the series $\frac1{n^n}$ also diverges. However by Cauchy root test the limit of $\frac1{n^n}$ appears to be $0<1$ which suggests the convergence of the series

• A series of nonnegative terms with all terms greater than or equal to a divergent series diverges. To dramatize your error, consider the series with all terms equal to zero. By your logic, since $0 < 1/n$ for all $n$, it would diverge. – quasi Oct 2 '17 at 9:46
• Your argument would be valid if $1/n^n \color{red}{>} 1/n$. But it is not. – M. Winter Oct 2 '17 at 9:50
• Series $\dfrac{1}{n^n}$ converges so quickly that just adding $5$ terms gives the result with $4$ exact decimals. Adding $10$ terms gives $1.29128\,599706$ which has eleven exact decimals – Raffaele Oct 2 '17 at 10:06
• @Raffaele: quite right, anyway shouldn't leave the OP the impression that this is a proof. What about $1/n^{n(20-n)}$ ? – Yves Daoust Oct 2 '17 at 10:14

We cannot conclude that since $\frac{1}{n^n} < \frac{1}{n}$ then $\sum_{n=1}^\infty \frac{1}{n^n}$ diverges.

If $\frac{1}{n^n} > \frac{1}{n}$ (which is not true), then we can conclude that $\sum_{n=1}^\infty \frac{1}{n^n}$ diverges.

You have used Cauchy root test to conclude that it converges.

If we want to use comparison test, notice that $$\frac{1}{n^n} \leq\frac{1}{n^2}$$

and since $\sum_{n=1}^\infty \frac1{n^2}$ conveges, hence $\sum_{n=1}^\infty \frac1{n^n}$ converges.

• Then is it that this given series converges by Cauchy root test ? – Nabendra Oct 2 '17 at 9:50
• yes, you have proven it using the root test. – Siong Thye Goh Oct 2 '17 at 9:51
• Now suppose if I use Ratio Test then, let u(n)=1/n^n and so u(n+1)=1/(n+1)^(n+1). Hence u(n)/u(n+1)=(1+1/n)^n * (n+1) and thus taking limit n tends to infinity on the above I get infinity(not sure whether calculations are correct) , which stands as greater than 1. So can I claim , the series in convergent by Ratio Test. – Nabendra Oct 2 '17 at 9:56
• Usually for ratio test, we compute $\left| \frac{u(n+1)}{u(n)}\right|$ and show that it is less than $1$. – Siong Thye Goh Oct 2 '17 at 10:28

The comparison test tells you that if $$a_n> b_n$$ and $$\sum_{n}a_n$$ converges, then $$\sum_n b_n$$ also converges.

It does not say that if $\sum_{n} a_n$ diverges then $\sum_{n} b_n$ diverges. If it did, then because $\frac{1}{2^n} < 1$, you could conclude that $\sum \frac1{2^n}$ also diverges.

In fact, you would need the inequality reversed, so if $a_n<b_n$ and $\sum a_n$ diverges, then $\sum b_n$ also diverges.

So, your logic is incorrect.

In fact, the series $\sum\frac{1}{n^n}$ converges by the comparison test with $\frac{1}{2^n}$ or with $\frac{1}{n^2}$.

• Does it appear convergent or divergent ? How gonna I check it ? – Nabendra Oct 2 '17 at 9:57
• @Nabendra I really really dislike users that don't read my answers before they ask questions about them. – 5xum Oct 2 '17 at 9:57
• I didn't understood your answer , as if I use 1/n the it diverges by comparison test instead of 1/n^2 or 1/2^n – Nabendra Oct 2 '17 at 10:04
• @Nabendra You didn't understand my answer because you didn't bother reading it to the end. So I don't see why I should bother helping you any further. – 5xum Oct 2 '17 at 10:04
• I do apologise, for my deed – Nabendra Oct 2 '17 at 10:07

A series converges if it has a converging upper bound and it diverges if it has a diverging lower bound. In other cases, you cannot conclude.

In this particular case, you can for instance exploit

$$\frac1{n^n}\le \frac1{2^n}$$ for $n\ge2$. Actually, the sequence converges extremely quickly, super-exponentially.