How to estimate $10^{\frac{1}{2}} + 10^{\frac{1}{3}} + \ldots + 10^{\frac{1}{10}}$

Question

I heard an interesting interview question recently, which was as follows:

Estimate the value of $X = 10^{\frac{1}{2}} + 10^{\frac{1}{3}} + \ldots + 10^{\frac{1}{10}}$.
You have 30 seconds to Compute it.

My approach was as follows:

• $10^\frac{1}{2} = \sqrt{10} \approx 3$
• $10^\frac{1}{3} = \sqrt[3]{10} \approx 2$
• $10^\frac{1}{10} \approx 1$

And then I guessed that $10^\frac{1}{4 \ldots 9}$ would average to around $1.2$ (because I guessed that they would be more biased towards $1$ than $2$, as the series would decrease quickly).
This gave me my estimate as $X = 3 + 2 + 1 + 6 * 1.2 = 13.2$, which is somewhat close to the answer. However, this is in no way rigorous or really theoretically motivated. Can someone provide a strategy that is?

For reference, the actual answer is

$X \approx 15.421$
$X \approx 15.421176223667882$ (with Calculators)

Answer 1

DISCLAIMER : When we come across such questions with time limit like 30 seconds , we take quite a while to think up some answer. Posting that end-result here will necessary take more than 30 seconds.
Objective of such questions is not the numerical answer , but the Thought Process.

Here is my thinking with my way to look at the Issue :

Compare "arithmetic mean" & "geometric mean" :
$X/9=(10^{\frac{1}{2}} + 10^{\frac{1}{3}} + \ldots + 10^{\frac{1}{10}})/9 \ge \sqrt[9]{( 10^{\frac{1}{2}} \times 10^{\frac{1}{3}} \times \ldots \times 10^{\frac{1}{10}})}$

$X/9 \ge \sqrt[9]{( 10^{\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{10}} )}$

Harmonic Numbers (Sum of reciprocals) is well-known & we get the Exponent $\approx 1.8$

$X/9 \ge \sqrt[9]{( 10^{1.8} )}$
$X \ge 9 \times 10^{1.8/9}$
$X \ge 9 \times 10^{0.2}$
$X \approx 10 \times 10^{0.2}$
$X \approx 10^{1.2}$

Here we have used the lower bound.
We can make the argument more rigorous by getting the upper bound by additionally using "quadratic mean" with "arithmetic mean" & "geometric mean" :
[[ https://en.wikipedia.org/wiki/QM-AM-GM-HM_inequalities ]]

30 SECOND ANSWER : use the fact that "arithmetic mean" is between "quadratic mean" & "geometric mean"

I think that Statement/Conclusion/Method is the Objective , not the numerical answer.

Answer 2

A quick and dirty approach is to note:

$$3 < 10^{1/2} < 4$$ $$2 < 10^{1/3} < 3$$ $$1 < 10^{1/k} < 2, k = 4...10$$

And thus, the sum must be somewhere between $3 + 2 + 7 \times 1 = 12$ and $4 + 3 + 7 \times 2 = 21$.

If you take the arithmetic mean of $12$ and $21$, you get $16.5$.

If you take the geometric mean, you get $2\sqrt{63} \approx 15.874508$. If you don't have a calculator, just note that $\sqrt{63}$ is slightly less than 8.

If you take the harmonic mean, you get $\frac{168}{11} = 15.272727...$

Answer 3

Here's one possible approach, letting $f(x)=10^{1/x}$:

• The sum of those ten numbers is $10$ times the average value of $f(1),\dots,f(10)$, which in turn is approximately $10$ times the average value of $f(x)$ on the interval $[1,10]$.
• The average value of $f(x)$ on the interval $[1,10]$ is larger than its value at the center $f(5.5)$ of the interval, since $f(x)$ is convex. So maybe the average value is about $f(5)$ (which is a little larger than $f(5.5)$ since $f(x)$ is decreasing).
• How big is $f(5) = 10^{1/5}$? Well, $1.6^5 = 16^5/10^5 = 2^{20}/10^5 \approx 10^6/10^5 = 10$. So maybe $10^{1/5} \approx 1.6$.

Therefore $10\cdot1.6=16$ is a reasonable approximation.

Answer 4

$$\sum_{j=2}^n n^{1\over j}\approx 1.3n\sqrt[n]{n}-0.84$$ This should be a good approximation

Answer 5

Very quick explanation

(Note: this was my reasoning when I tried to solve it in 30 seconds. Don't take it too seriously)

• Each number $>1$ so the answer $>9$.
• $10^{1/2}>3$ and $10^{1/3}>2$ so we add $3$ to the answer.
• For $10^{1/4,5}$ we add $1$, for $10^{1/6,7,8}$ we add $1$, for $10^{1/9,10}$ it is slightly less than $1$ but in the last line we have underestimated, so we still add $1$.
• So the answer is around $15$.

Slightly more justified explanation

• Using $(1+1/k)^k\approx e$ and $10>e^2$, we approximate that $10^{1/10}>1.2$ (and probably closer to $1.3$).
• Linearly interpolating between $10^{1/10}\approx 1.2$ and $10^{1/4}\approx 1.8$ (just because it's slightly less than $2$) makes $10.5$. This is an overestimate because $1/x$ is convex.
• So we compensate by assuming $10^{1/3}\approx 2$ and $10^1/2\approx 3$.
• Making $15.5$.

Answer 6

Just to throw out a wholly unsophisticated answer based strictly on rough guessing, expanding on my comment to the question: I already know $\sqrt{10} \approx 3.162$, but I'll just pick $3.2$. Then $\sqrt[4]{10} \approx 1.8$ and $\sqrt[8]{10} \approx 1.3$.

Then I'll estimate $\sqrt[3]{2} \approx 2.2$ which leads to $\sqrt[6]{10} \approx 1.5$.

That gives me enough of a scaffold to interpolate $\sqrt[5]{10} \approx 1.6, \sqrt[7]{10} \approx 1.4$, and $\sqrt[9]{10} \approx 1.2 \approx \sqrt[10]{10}$. I have a sneaking suspicion the early ones are overestimates and the late ones are underestimates, but that's OK; I'm only worried about their sum:

$$ \sum_{k=2}^{10} \sqrt[k]{10} \approx 3.2 + 2.2 + 1.8 + 1.6 + 1.5 + 1.4 + 1.3 + 1.2 + 1.2 = 15.4 $$

That's pretty close to the actual answer. On another day, maybe I estimate higher for some and lower for others, but I doubt I'll be off by more than $0.3$ or so.