Limit log-sum of exponentials

Question

I'm trying to compute the following limit:

$$\lim_{\lambda \rightarrow \infty} \frac{1}{\lambda}\log\sum_{i=1}^n \exp[\lambda a_i]$$

I tried to solve with L'hoptials: $$= \lim_{\lambda \rightarrow \infty} \frac{\frac{d}{d\lambda}\log\sum_{i=1}^n \exp[\lambda a_i]}{\frac{d}{d\lambda}\lambda}$$

$$ = \lim_{\lambda \rightarrow \infty} \frac{\sum_{i=1}^n a_i\exp[\lambda a_i]}{\sum_{i=1}^n \exp[\lambda a_i]}$$

Now here is where I'm stuck. Is there any way to reduce the quotient to solve the limit? Applying L'Hopitals again doesn't seem to buy you anything.

Thanks!

Answer 1

The answer should be the largest of the $a_i$. To get to it intuitively from what you have in your last equation, note that if $a>b$, $e^{\lambda a}$ grows faster than $e^{\lambda b}$, since $e^{\lambda(a-b)} \to \infty$. You can use this principle to say that the top is dominated by the term where $a_i$ is largest, as is the bottom, so the limit tends to ratio of the coefficients of these terms, which is $a_i$.

The other way is to look at it as $$ \log{\left( \lim_{\lambda \to \infty} \left( \sum_{i=1}^n (e^{a_i})^{\lambda} \right)^{1/\lambda} \right)} = \log{\left( \lim_{\lambda \to \infty} \lVert (e^{a_1}, \dots, e^{a_n} ) \rVert_{\lambda} \right)}, $$ which you can since the logarithm is continuous.

You can then factor out the largest e^{a_i} using the linearity of the norm, and the rest is a vector with one $1$ and the other numbers $\leq 1$. The internal sum is then bounded above by $n$, and $n^{1/\lambda} \to 1$, so the answer is $\log{e^{a_i}}=a_i$.

Answer 2

Whatever could be the $a_i$'s, you could sort them by increasing values and rewrite $$S=\sum_{i=1}^n e^{\lambda a_i}=\sum_{i=1}^n e^{\lambda b_i}$$ in which $b_1\leq b_2 \leq b_3\leq \cdots\leq b_n$. So, $$S=\sum_{i=1}^n e^{\lambda b_i}=e^{\lambda b_n}\sum_{i=1}^ne^{\lambda (b_i-b_n)}=e^{\lambda b_n}\Big(1+\sum_{i=1}^{n-1}e^{\lambda (b_i-b_n)}\Big)$$ So $$\log(S)=\lambda b_n+\log\Big(1+\sum_{i=1}^{n-1}e^{\lambda (b_i-b_n)}\Big)$$ $$\frac {1}\lambda \log(S)=b_n+\frac {1}\lambda \log\Big(1+\sum_{i=1}^{n-1}e^{\lambda (b_i-b_n)}\Big)$$ and in the summation just appear small numbers. Then, the result Chappers gave.

Answer 3

As log is a monotone increasing function, and the sum inside contains only non-negative terms, we may use the squeeze theorem, by replacing or dropping any of terms.

For the above direction: $$ \lim_{\lambda \rightarrow\infty}{\frac 1 \lambda \log({\sum_{i=1}^n{\exp[\lambda a_i]}})} \le \lim_{\lambda \rightarrow\infty}{\frac 1 \lambda \log({n\cdot\exp[\lambda \max a_i]})} = $$ $$ \lim_{\lambda \rightarrow\infty}{\frac 1 \lambda \log(n)} + \lim_{\lambda \rightarrow\infty}{\frac 1 \lambda \log({\exp[\lambda \max a_i]})} = 0 + \max a_i $$

For the below direction, proceed similarly only this time drop all the terms in the sum except $e^{\max {x_i}}$.