Show that the following infinite series ∑i=1∞f(n)n2

Question

Show that the following infinite series  ∑i=1∞f(n)n2

fumanchuecf · Accepted Answer

The goal is to use the rearrangement inequality to compare the series to the harmonic series

\sum_{n = 1}^{\infty} \frac{1}{n}

which diverges. Consider a partial sum

S (N) = \sum_{n = 1}^{N} \frac{f (n)}{n^{2}}

Let

x_{1}, \dots, x_{N}

be the values of

f (n)

for

1 \leq n \leq N

, labeled such that

x_{1} \leq \dots \leq x_{N}

Let

y_{j} = \frac{1}{{(N - j + 1)}^{2}}

for

1 \leq j \leq N

. The rearrangement inequality tells us that

x_{N} y_{1} + \dots + x_{1} y_{N} \leq x_{σ (1)} y_{1} + \dots + x_{σ (N)} y_{N}

(1)
for any permutation

σ

of

1, \dots, M = N

. Choose

σ

such that

x_{σ (j)} = f (N - j + 1)

so that the right-hand side of (1) is equal to

\frac{f (1)}{1} + \frac{f (2)}{4} + \dots + \frac{f (N)}{N^{2}} = S (N)

Now we examine the left-hand side of (1). Since f is injective and

x_{N}

is the largest of the

x_{i}

, it must be that

x_{N} \geq N

. By induction it follows that

x_{j} \geq j

for

1 \leq j \leq N

. Therefore

1 + \frac{1}{2} + \dots + \frac{1}{N} \leq x_{N} y_{1} + \dots + x_{1} y_{n}

We thus have

1 + \frac{1}{2} + \dots + \frac{1}{N} \geq S (N)

hence your series diverges by the comparison test.

izumrledk · Accepted Answer

The rearrangement inequality says the partial sum  ∑n=1Mf(n)n2≥∑n=1Mbnn2  where b1,…,bM are f(1),…,f(M) rearranged into increasing order. Since bn≥n, we get  ∑n=1Mf(n)n2≥∑n=1Mnn2=∑n=1M1n  Now what do you know about the sum on the right?

RizerMix · Accepted Answer

Here is a direct proof without the use of the rearrangement inequality. For every N, you have ∑n=N+13Nf(n)n2≥∑n=N+13Nf(n)(3N)2≥N29N2 because the f(n) for n∈[N+1,3N] are 2N distinct integers, so at least N of them are greater than N. So ∑N+13Nf(n)n2≥19 so it diverges.

Show that the following infinite series \sum_{i=1}^\infty\frac{f(n)}{n^2}

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