A Riemannian manifold $(M,g)$ is said to have flat ends if the curvature tensor of $g$ vanishes outside a compact set $K$. I was wondering if such manifolds are of bounded geometry. Recall that a manifold is of bounded geometry if

 1. The curvature tensor and all its covariant derivatives are uniformly bounded.
 2. The injectivity radius has a uniform positive lower bound.

It is obvious that a manifold with flat ends satisfies the first property, but it is not clear to me that a manifold with flat ends satisfies the second property. Two simple counterexamples come to mind. 

 1. The manifold $M=\mathbb{R}^n-\{0\}$ has flat ends, but has no uniform lower bound on the injectivity radius.
 2. A cylinder $\mathbb{R}\times S_r^1$ of radius $r$ has injectivity radius $\pi r$. Take a countable union of such cylinders with decreasing radius
$$M=\coprod_{n=1}^\infty \mathbb{R}\times S_{\frac 1n}^1.$$
This manifold is flat, but does not admit a uniform positive lower bound on the injectivity radius.

Counterexample one might be excluded by assuming that $g$ is complete, and counterexample 2 might be excluded by demanding that $M$ is connected. Therefore my question is:

<blockquote> Is any connected and complete Riemannian manifold $(M,g)$ with flat ends of bounded  geometry? </blockquote>