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

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)

• This is an interesting question because the number $10$ is too small for limit heuristics to be used (see here for what one can do if $10$ was replaced by say $100$). Then again, I don't see how one can get three significant decimal places in $30$ seconds, unless one is allowed to use a calculator. Commented Apr 2 at 6:03
• A bit in line with your approach: take $10^\frac{1}{2}$ separately, then you have 8 values decreasing from 2 to 1 left (mean 1.5), so 3 + 8*1.5 = 15 (all approximately...). This does not work well however if you add more terms on the right side. Commented Apr 2 at 6:46
• To be honest, in $30$ seconds, I found it easier to just estimate each of these to the nearest $0.1$ or so. There are few enough and I can estimate square roots pretty well, plus the cube root of $10$, and interpolate the rest. I ended up getting $3.2 + 2.2 + 1.8 + 1.6 + 1.5 + 1.4 + 1.3 + 1.2 + 1.2 = 15.4$. With more terms on the right, I might try something to estimate the tail, but it's pretty well behaved out there. Commented Apr 3 at 17:53
• I ended up expanding the above into an actual answer, for the sake of posterity. :-D Commented Apr 3 at 18:05

## 6 Answers

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.

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 $$\frac{168}{11} = 15.272727...$$

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.

• There are $9$ numbers in the sum $X$.
– Gary
Commented Apr 2 at 9:25

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

• It works well, at least for $n = 10$, but how would one derive that expression and those constants in $30$ seconds? Commented Apr 2 at 23:36
• Nice. This is 15.52 which is close to the actual 15.42. But how did you come up with it? And how would you compute it quickly? Commented Apr 3 at 0:04
• For large $n$, the sum is asymptotic to $n$.
– Gary
Commented Apr 3 at 0:05
• @BenjaminWang this comes from the integral approximation of the sum, atleast the part $n\sqrt[n]{n}$, but it involved the exponential integral which I simplified to $0.84$ and $1.3$ (approximately)
– Masd
Commented Apr 3 at 0:48
• @BrianTung if you picture the integral approximation you can get the $n\sqrt[n]n$ part, the constants would be harder
– Masd
Commented Apr 3 at 0:52

## 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 $$15.5$$.

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.

• Taking about $10$ times longer to try to do it to two digits only got me to $15.44$, so really no more accurate than before. Commented Apr 3 at 18:15