Prove $\exp\left(\sum_{s = 1}^t \frac{1}{s}\right) \geq 1 + t$

Question

I am in the middle of a long proof, part of which requires that

\begin{align*} \exp\left(\sum_{s = 1}^t \frac{1}{s}\right) \geq 1 + t. \end{align*}

Plotting on Desmos, this appears to be true:

I would appreciate any help in proving this. I don't think I can use that $\exp(x) \geq x + 1$, and converting the exponential of the sum into the product of exponentials doesn't obviously lead anywhere.

I appreciate any help.

Answer 1

HINT [click below to unveil full answer]: $$\ln(1+t) \ = \ \int_{1}^{1+t}\frac{dx}{x}$$ $$=\sum_{s=1}^t \int_s^{s+1} \frac{dx}{x} \ \le \ \sum_{s=1}^t \frac{1}{s},$$

the last inequality following from $\frac{1}{x}$ being a decreasing function in $x$ so $\int_s^{s+1}\frac{dx}{x}$ is upper-bounded by $\frac{1}{s} ×((s+1)-s)$ $=$ $\frac{1}{s}×1$ $=\frac{1}{s}$.

So in particular then,

$$\ln(1+t) \ \le \ \sum_{s=1}^t \frac{1}{s}.$$ What do you observe taking the exponential of both sides of the above inequality.

Answer 2

We have $$\exp(1/s)\ge 1+{1\over s}={s+1\over s}$$ Hence $$\exp\left (\sum_{s=1}^t{1\over s}\right )=\prod_{s=1}^t\exp(1/s)\\ \ge \prod_{s=1}^t{s+1\over s}=t+1$$