Question

Let $a_k = \frac{k}{(k-1)^{4/3} + k^{4/3} + (k+1)^{4/3}}$ and $S_n = \sum_{k=1}^{n} a_k$, then

• A. $S_{26} > \frac{17}{4}$
• B. $S_{26} < \frac{17}{4}$
• C. $S_{999} < 50$
• D. $S_{999} > 50$

Answer 1

Answer: (B), (C)

Options (B) and (C)

Answer 2

We wish to study

$$a_k=\frac{k}{(k-1)^{4/3}+k^{4/3}+(k+1)^{4/3}}$$

and then

$$S_n=\sum_{k=1}^n a_k.$$

A key observation is that for large $k$ the dominant behavior is given by the powers of $k$. Notice that

$$(k-1)^{4/3}\sim k^{4/3},\quad k^{4/3}\sim k^{4/3},\quad (k+1)^{4/3}\sim k^{4/3}.$$

In fact, the denominator behaves approximately as

$$(k-1)^{4/3}+k^{4/3}+(k+1)^{4/3}\approx 3k^{4/3}\quad\text{for large }k.$$

Thus for large $k$ we have

$$a_k\approx\frac{k}{3k^{4/3}}=\frac{1}{3\,k^{1/3}}.$$

Once you have the asymptotic approximation

$$S_n\approx\sum_{k=1}^n \frac{1}{3\,k^{1/3}},$$

it is natural to use an integral to estimate the sum. In particular,

$$\sum_{k=1}^{n} \frac{1}{3\,k^{1/3}}\approx \frac{1}{3}\int_{1}^{n}x^{-1/3}\,dx.$$

Since

$$\int x^{-1/3}\,dx=\frac{3}{2}x^{2/3}+C,$$

we get

$$S_n\approx \frac{1}{3}\cdot\frac{3}{2}\Bigl(n^{2/3}-1\Bigr)=\frac{n^{2/3}-1}{2}.$$

Now let’s apply this to the two cases:

1. For $n=26$:

We estimate

$$S_{26}\approx\frac{26^{2/3}-1}{2}.$$

Since $27^{1/3}=3$ and $26^{1/3}$ is slightly less (approximately $2.96$), we have

$$26^{2/3}\approx(2.96)^2\approx 8.77.$$

Then,

$$S_{26}\approx\frac{8.77-1}{2}=\frac{7.77}{2}\approx3.885.$$

Now, $\frac{17}{4}=4.25$. Thus $S_{26} \approx 3.89<4.25$. Therefore option (B) is true.

2. For $n=999$:

Similarly,

$$S_{999}\approx\frac{999^{2/3}-1}{2}.$$

Note that $1000^{1/3}=10$ so $999^{1/3}$ is very nearly 10 and hence $999^{2/3}$ is nearly $10^2=100$. In fact, one may estimate

$$S_{999}\approx\frac{100-1}{2}=\frac{99}{2}=49.5.$$

This shows that $S_{999}<50$ so option (C) is true.

Thus our conclusions are:

• (B) $S_{26}<\frac{17}{4}$
• (C) $S_{999}<50$

Minimal Core Explanation

1. Asymptotic behavior: For large $k$, approximate $$a_k \sim \frac{1}{3\,k^{1/3}}.$$
2. Integral estimate: $$S_n\approx\frac{1}{3}\int_1^n x^{-1/3}\,dx=\frac{n^{2/3}-1}{2}.$$
3. Apply for $n=26$: $\frac{26^{2/3}-1}{2}\approx 3.89 < 4.25=\frac{17}{4}$.
4. Apply for $n=999$: $\frac{999^{2/3}-1}{2}\approx 49.5 < 50$.